CHEM 122 - GENERAL CHEMISTRY
II - Precipitation calculations
Question Bank - Set 1
Liberty University
Question 1
Question
Calculate the concentration of sulfate ions in a solution when 50.0 mL of a 0.10
M solution of barium chloride is mixed with excess sulfuric acid and the resulting
precipitate is filtered off. The volume of the filtrate is 75.0 mL. (Given: Molar
mass of BaCl2= 208.23 g/mol, molar mass of BaSO4= 233.39 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sulfuric acid. Step 2: Calculate the moles of barium chloride used
in the reaction. Step 3: Use the stoichiometry of the balanced chemical equation
to find the moles of sulfate ions produced. Step 4: Calculate the concentration
of sulfate ions in the solution.
Step 1: The balanced chemical equation for the reaction is:
BaCl2+ H2SO4→BaSO4+ 2HCl
Step 2: Number of moles of BaCl2= concentration ×volume (in L) Number
of moles of BaCl2= 0.10 mol/L ×50.0 mL ×(1 L / 1000 mL) = 0.005 mol
Step 3: From the balanced chemical equation, it is clear that 1 mole of
BaCl2reacts with 1 mole of BaSO4. Therefore, the moles of BaSO4formed is
also 0.005 mol.
Step 4: Since the volume of the solution after filtration is 75.0 mL, the total
volume of the solution is 125.0 mL or 0.125 L. Concentration of sulfate ions =
moles/volume = 0.005 mol / 0.125 L = 0.04 M
Therefore, the concentration of sulfate ions in the solution is 0.04 M.
Question 2
Question
A solution contains 250 mL of 0.2 M silver nitrate (AgNO3) and 300 mL of 0.3
M sodium chloride (NaCl). Calculate the mass of silver chloride (AgCl) that
will precipitate when the two solutions are mixed.
(Hint: AgCl precipitates according to the following reaction: AgNO3+
NaCl →AgCl ↓+NaNO3)
Solution
Step 1: Calculate the moles of AgNO3and NaCl using the formula n= M ×V.
For AgNO3:
nAgNO3= 0.2 mol/L ×0.250 L = 0.05 mol
For NaCl:
nNaCl = 0.3 mol/L ×0.300 L = 0.09 mol
Step 2: Determine the limiting reactant by comparing the moles of each
reactant. According to the balanced chemical equation, 1 mole of AgNO3reacts
with 1 mole of NaCl to form 1 mole of AgCl. Since the mole ratio is 1:1, the
limiting reactant is the one that produces the least amount of product. In this
case, AgNO3is the limiting reactant because it produces 0.05 moles of AgCl
compared to 0.09 moles from NaCl.
Step 3: Calculate the mass of AgCl precipitated using the molar mass of
AgCl (143.32 g/mol).
Mass of AgCl = nAgCl×Molar mass of AgCl = 0.05 mol×143.32 g/mol = 7.166 g
Therefore, when 250 mL of 0.2 M AgNO3is added to 300 mL of 0.3 M NaCl,
7.166 g of AgCl will precipitate.
Question 3
Question
Calculate the solubility of lead(II) iodide (PbI2) at 25
°
C. The Ksp of PbI2is
7.1×10−9.
Solution
Step 1: Write the solubility equilibrium expression for lead(II) iodide:
PbI2⇌Pb2+ + 2I−
The equilibrium constant expression for this reaction is:
Ksp = [Pb2+][I−]2
2
Step 2: Let’s assume that the solubility of PbI2is x. This means the equi-
librium concentrations of Pb2+ and I−will be xand 2xrespectively.
Step 3: Substitute the equilibrium concentrations into the Ksp expression:
7.1×10−9= (x)(2x)2
Step 4: Solve for x:
7.1×10−9= 4x3
x3=7.1×10−9
4
x=3
r7.1×10−9
4
Step 5: Calculate the solubility of lead(II) iodide:
x≈5.72 ×10−3M
Therefore, the solubility of lead(II) iodide at 25
°
C is approximately 5.72 ×
10−3M.
Question 4
Question
Calculate the molarity of barium chloride (BaCl2) solution when 50.0 mL of a
0.200 M solution of silver nitrate (AgNO3) is added to 75.0 mL of a solution
containing 0.150 M of sodium sulfate (Na2SO4). Assume the reaction goes to
completion and that barium sulfate (BaSO4) precipitates.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium sulfate to find the net ionic equation.
AgNO3+ Na2SO4→Ag2SO4+ NaNO3
The net ionic equation is:
Ag++ SO2−
4→Ag2SO4
Step 2: Determine the limiting reagent to find the moles of barium sulfate
formed.
From the net ionic equation, it is clear that 1 mole of barium sulfate is
formed from 1 mole of silver nitrate.
Moles of silver nitrate = 0.0500 L ×0.200 mol/L = 0.0100 mol
Since the reaction ratio is 1:1, moles of barium sulfate formed = 0.0100 mol
Step 3: Calculate the moles of barium chloride formed using the mole ratio
from the balanced equation.
From the balanced equation for the precipitation reaction:
BaCl2+ Na2SO4→BaSO4+ NaCl
3
We have a 1:1 mole ratio between barium chloride and barium sulfate.
Therefore, moles of barium chloride = moles of barium sulfate = 0.0100 mol
Step 4: Calculate the molarity of the barium chloride solution.
Volume of barium chloride solution = 75.0 mL + 50.0 mL = 125.0 mL =
0.1250 L
Molarity of barium chloride solution = 0.0100 mol
0.1250 L = 0.0800 M
Therefore, the molarity of the barium chloride solution is 0.0800 M.
Question 5
Question
In a lab experiment, a solution of 50 mL of 0.2 M potassium iodide (KI) is mixed
with 30 mL of 0.3 M lead(II) nitrate (Pb(NO3)2) solution. The only product
formed is lead(II) iodide (PbI2), which precipitates out of the solution.
Calculate the maximum mass of lead(II) iodide that can be formed in grams.
(Hint: The balanced chemical equation for the reaction is Pb(NO3)2+2KI →
PbI2+ 2KNO3)
Solution
Step 1: Calculate the moles of potassium iodide (KI) and lead(II) nitrate
(Pb(NO3)2) used in the reaction. Given: For potassium iodide (KI): Volume =
50 mL = 0.05 L Molarity = 0.2 M
Number of moles of KI = Molarity ×Volume Number of moles of KI =
0.2 mol/L ×0.05 L = 0.01 mol
For lead(II) nitrate (Pb(NO3)2): Volume = 30 mL = 0.03 L Molarity = 0.3
M
Number of moles of Pb(NO3)2= Molarity ×Volume Number of moles of
Pb(NO3)2= 0.3 mol/L ×0.03 L = 0.009 mol
Step 2: Determine the limiting reactant. From the balanced chemical equa-
tion, 1 mole of Pb(NO3)2reacts with 2 moles of KI to form 1 mole of PbI2.
Since the ratio is 1:2, Pb(NO3)2is the limiting reactant because it produces
fewer moles of PbI2compared to KI.
Step 3: Calculate the maximum mass of lead(II) iodide (PbI2) that can be
formed. From the stoichiometry of the reaction, 1 mole of PbI2has a molar
mass of 461 g/mol. Number of moles of PbI2= 0.009 mol (since Pb(NO3)2is
the limiting reactant)
Maximum mass of PbI2= Number of moles ×Molar mass Maximum mass
of PbI2= 0.009 mol ×461 g/mol = 4.149 g
Therefore, the maximum mass of lead(II) iodide that can be formed is 4.149
grams.
4
Question 6
Question
Calculate the solubility of silver chloride (AgCl) in a 0.010 M solution of silver
nitrate (AgNO3). The Ksp of AgCl is 1.8×10−10.
Solution
Step 1: Write the balanced equation for the dissociation of AgCl:
AgCl ⇌Ag++ Cl−
Step 2: Set up an ICE (Initial, Change, Equilibrium) table for the dissocia-
tion of AgCl:
Substance AgCl Ag+Cl−
Initial (M) x0 0
Change (M) −x+x+x
Equilibrium (M) x−x x x
Step 3: Write the expression for the solubility product Ksp:
Ksp = [Ag+][Cl−]
Step 4: Substitute the equilibrium concentrations into the Ksp expression:
1.8×10−10 =x×x
Step 5: Solve for xto find the equilibrium concentration of Ag+and Cl−
ions:
x=p1.8×10−10 = 1.34 ×10−5
Step 6: The solubility of AgCl in the 0.010 M solution of AgNO3is equal to
the concentration of Ag+ions, which is 1.34 ×10−5M.
Question 7
Question
Calculate the mass of lead(II) iodide that will precipitate when 50.0 mL of 0.200
M lead(II) nitrate is mixed with 25.0 mL of 0.300 M potassium iodide. Assume
the reaction goes to completion.
Solution
Step 1: Write down the balanced chemical equation for the reaction.
Pb(NO3)2(aq) + 2KI (aq) →PbI2(s) + 2KNO3(aq)
5
Step 2: Determine the limiting reactant. Calculate the moles of lead(II)
nitrate and potassium iodide: For lead(II) nitrate: (0.200 M) ×(0.0500 L) =
0.0100 mol For potassium iodide: (0.300 M) ×(0.0250 L) = 0.00750 mol
Since potassium iodide is limiting (less moles), it will completely react.
Step 3: Calculate the moles of lead(II) iodide precipitated. Using stoichiom-
etry from the balanced equation: 1 mol of Pb(NO3)2produces 1 mol of PbI2
Therefore, 0.00750 mol of PbI2will be formed.
Step 4: Convert moles of lead(II) iodide to mass. Given molar mass of PbI2
is approximately 461.01 g/mol, Mass of PbI2= 0.00750 mol ×461.01 g/mol =
3.46 g
So, the mass of lead(II) iodide that will precipitate is 3.46 grams.
Question 8
Question
A researcher is studying the precipitation patterns in a region and collects the
following data for the month of April:
Total rainfall: 120 mm
Number of rainy days: 15
Assuming that the rainfall is evenly distributed over the rainy days, deter-
mine the average precipitation intensity in mm/day during the month of April.
Solution
To find the average precipitation intensity in mm/day during the month of April,
we need to divide the total rainfall by the number of rainy days.
Step 1: Calculate the average precipitation intensity.
Average precipitation intensity = Total rainfall
Number of rainy days
Average precipitation intensity = 120 mm
15 = 8 mm/day
Step 2: Answer: The average precipitation intensity during the month of
April is 8 mm/day.
Question 9
Question
A chemist needs to prepare 500 mL of a 0.2 M lead nitrate (Pb(NO3)2) solution
by dissolving solid lead nitrate in water. If the molar mass of lead nitrate is
331.21 g/mol, determine the mass of lead nitrate needed.
6
Solution
Step 1: Calculate the number of moles of lead nitrate needed. We can use the
formula M=n
V, where Mis the molarity, nis the number of moles, and Vis
the volume in liters.
Given: M= 0.2 M V= 500 mL = 0.5 L
Plugging in the values: n=M×V= 0.2 mol/L ×0.5 L = 0.1 mol
Therefore, the chemist needs 0.1 moles of lead nitrate.
Step 2: Calculate the mass of lead nitrate needed. We can use the formula
m=n×MM, where mis the mass, nis the number of moles, and MM is the
molar mass.
Given: n= 0.1 mol MM = 331.21 g/mol
Plugging in the values: m= 0.1 mol ×331.21 g/mol = 33.121 g
Therefore, the chemist needs 33.121 grams of lead nitrate to prepare 500 mL
of a 0.2 M solution.
Question 10
Question
A chemistry student wants to determine the concentration of chloride ions in a
solution. To do this, they add silver nitrate (AgNO3) to a 100.0 mL sample of
the solution, causing a white precipitate of silver chloride (AgCl) to form. The
student then filters the precipitate, dries it, and weighs it. The mass of the dry
silver chloride obtained is 0.345 g. Calculate the concentration of chloride ions
in the original solution in units of mol/L. The molar mass of AgCl is 143.32
g/mol.
Solution
Step 1: Write the balanced chemical equation for the reaction that occurs when
silver nitrate is added to chloride ions in solution. The balanced chemical equa-
tion is:
Ag++ Cl−→AgCl
Step 2: Calculate the moles of AgCl formed. Given: Mass of AgCl = 0.345
g Molar mass of AgCl = 143.32 g/mol
Moles of AgCl = Mass
Molar mass =0.345 g
143.32 g/mol ≈0.0024 mol
Step 3: Calculate the moles of chloride ions in the original solution. Since 1
mole of AgCl is produced when 1 mole of chloride ion is present, the moles of
chloride ions in the original solution is also 0.0024 mol.
Step 4: Determine the volume of the original solution. Given: Volume of
solution = 100.0 mL = 0.100 L
7
Step 5: Calculate the concentration of chloride ions in the original solution.
The concentration of chloride ions is given by:
Concentration = Moles of solute
Volume of solution (in L) =0.0024 mol
0.100 L = 0.024 mol/L
Answer: The concentration of chloride ions in the original solution is 0.024
mol/L.
Question 11
Question
Calculate the concentration of a new precipitate formed when 50.0 mL of a 0.200
M solution of calcium chloride is mixed with 100.0 mL of a 0.300 M solution of
sodium carbonate. Assume the reaction goes to completion and that the volume
of the final solution is 150.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride (CaCl2) and sodium carbonate (Na2CO3):
CaCl2+ Na2CO3→CaCO3+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant: For calcium chloride: Number of moles = Concentration ×Volume
= 0.200 M ×0.0500 L = 0.0100 mol For sodium carbonate: Number of moles =
Concentration ×Volume = 0.300 M ×0.100 L = 0.0300 mol
Since calcium chloride produces 1 mol of precipitate for every 1 mol of sodium
carbonate, calcium chloride is the limiting reactant.
Step 3: Calculate the number of moles of precipitate formed (calcium car-
bonate): Number of moles of CaCO3= Number of moles of limiting reactant
= 0.0100 mol
Step 4: Calculate the concentration of the precipitate in the final solution:
Total volume of final solution = 150.0 mL = 0.150 L Concentration of CaCO3
=0.0100 mol
0.150 L Concentration of precipitate = 0.067 M
Therefore, the concentration of the new precipitate formed in the final solu-
tion is 0.067 M.
Question 12
Question
Calculate the mass of silver chloride (AgCl) that will precipitate when a so-
lution containing 0.050 moles of silver nitrate (AgNO3) is added to a solution
containing 0.035 moles of sodium chloride (NaCl). Assume that the reaction
goes to completion.
8
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant to find the maximum amount of
silver chloride that can be formed. First, calculate the theoretical yield of AgCl
using the moles of AgNO3:
moles of AgCl = moles of AgNO3= 0.050 mol
Step 3: Calculate the theoretical yield of AgCl using the moles of NaCl:
moles of AgCl = moles of NaCl = 0.035 mol
Step 4: Identify the limiting reactant by comparing the amounts of AgCl
produced from each reactant. Since AgNO3produces more AgCl (0.050 moles)
than NaCl (0.035 moles), NaCl is the limiting reactant.
Step 5: Calculate the mass of silver chloride precipitated from the limiting
reactant (sodium chloride):
molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
mass of AgCl = moles of AgCl ×molar mass of AgCl
mass of AgCl = 0.035 mol ×143.32 g/mol = 5.0132 g
Thus, the mass of silver chloride that will precipitate when 0.050 moles of
silver nitrate is added to a solution containing 0.035 moles of sodium chloride
is 5.0132 grams.
Question 13
Question
A solution is prepared by mixing 100 mL of a 0.2 M calcium chloride solution
with 200 mL of a 0.4 M sodium sulfate solution. Determine if a precipitate will
form when these two solutions are mixed.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride (CaCl2) and sodium sulfate (Na2SO4).
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Calculate the initial moles of each ion in the solutions. For calcium
chloride solution: - Moles of Ca2+ ions: 0.2 M ×0.1 L = 0.02 moles - Moles of
Cl−ions: 0.2 M ×0.1 L = 0.02 moles
9
For sodium sulfate solution: - Moles of Na+ions: 0.4 M ×0.2 L = 0.08 moles
- Moles of SO2−
4ions: 0.4 M ×0.2 L = 0.08 moles
Step 3: Determine the limiting reactant by examining the mole ratio in the
balanced equation. From the equation, the mole ratio of CaCl2to Na2SO4is
1:1. Therefore, both reactants are in stoichiometric amounts and there is no
limiting reactant.
Step 4: Write the net ionic equation for the precipitation reaction.
Ca2+ + SO2−
4→CaSO4
Step 5: Check the solubility rules to determine if CaSO4is insoluble. Cal-
cium sulfate (CaSO4) is insoluble according to most solubility rules, and thus a
precipitate will form when the two solutions are mixed.
Therefore, when the 0.2 M calcium chloride solution is mixed with the 0.4
M sodium sulfate solution, a white precipitate of calcium sulfate will form.
Question 14
Question
Calculate the mass of lead(II) chloride (P bCl2) that can be formed when 500.0
mL of 0.200 M lead(II) nitrate (P b(NO3)2) solution is reacted with excess
sodium chloride (NaCl). Assume the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and sodium chloride to form lead(II) chloride.
P b(NO3)2+ 2NaCl →P bCl2+ 2NaNO3
Step 2: Calculate the moles of lead(II) nitrate using the given volume and
molarity.
moles = volume ×molarity = 0.5 L ×0.200 M = 0.100 moles
Step 3: Use the mole ratio from the balanced equation to find the moles of
lead(II) chloride that can be formed.
moles of PbCl2= 0.100 moles ×1 mol PbCl2
1 mol Pb(NO3)2
= 0.100 moles
Step 4: Calculate the mass of lead(II) chloride formed using the molar mass
of P bCl2.
molar mass of PbCl2= 207.2 g/mol + 2 ×35.45 g/mol = 278.1 g/mol
mass of PbCl2= 0.100 moles ×278.1 g/mol = 27.81 g
Answer: The mass of lead(II) chloride that can be formed is 27.81 grams.
10
Question 15
Question
A solution contains 100 mL of 0.2 M calcium chloride and 200 mL of 0.15 M
sodium sulfate. What mass of calcium sulfate can be precipitated out?
(Hint: Use stoichiometry to calculate the limiting reactant and then deter-
mine the mass of the precipitate.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between calcium chloride (CaCl2) and sodium sulfate (Na2SO4) to form calcium
sulfate (CaSO4) and sodium chloride (NaCl).
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Calculate the moles of each reactant using the molarity formula
M=moles
liters . For calcium chloride:
0.2 M = molesCaCl2
0.1 L
molesCaCl2= 0.2 mol
For sodium sulfate:
0.15 M = molesNa2SO4
0.2 L
molesNa2SO4= 0.03 mol
Step 3: Determine the limiting reactant by comparing the moles of each
reactant and their stoichiometric coefficients. Since 1 mole of calcium chloride
reacts with 1 mole of sodium sulfate, whichever reactant has fewer moles is the
limiting reactant. In this case, sodium sulfate is the limiting reactant.
Step 4: Calculate the moles of calcium sulfate that can be formed based on
the limiting reactant (sodium sulfate). From the reaction, 1 mole of sodium
sulfate reacts to form 1 mole of calcium sulfate.
molesCaSO4= 0.03 mol
Step 5: Calculate the mass of calcium sulfate precipitated using the molar
mass of calcium sulfate (CaSO4) which is 136.14 g/mol.
MassCaSO4= molesCaSO4×Molar massCaSO4
MassCaSO4= 0.03 mol ×136.14 g/mol = 4.0842 g
Therefore, approximately 4.08 g of calcium sulfate can be precipitated out.
11
Question 16
Question
A university chemistry lab is conducting an experiment where a solution of
barium chloride (BaCl2) is mixed with a solution of sulfuric acid (H2SO4). If
100.0 mL of 0.200 M barium chloride solution is mixed with 50.0 mL of 0.500
M sulfuric acid solution, what mass of barium sulfate (BaSO4) will precipitate
out?
(Hint: The reaction is: BaCl2+ H2SO4→BaSO4+ 2HCl.)
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sulfuric acid.
BaCl2+ H2SO4→BaSO4+ 2HCl
Step 2: Calculate the moles of each reactant used in the reaction.
Moles of barium chloride:
Moles of BaCl2= Volume (L)×Molarity = 0.100 L×0.200 mol/L = 0.020 mol
Moles of sulfuric acid:
Moles of H2SO4= Volume (L)×Molarity = 0.050 L×0.500 mol/L = 0.025 mol
Step 3: Determine the limiting reactant. Since the reaction requires 1 mole
of BaCl2to react with 1 mole of H2SO4, and there are more moles of BaCl2
available, H2SO4is the limiting reactant.
Step 4: Calculate the mass of barium sulfate precipitated out using the
limiting reactant.
Moles of BaSO4= Moles of limiting reactant = 0.025 mol
Mass of BaSO4= Moles×Molar mass = 0.025 mol×(137.3 g/mol+32.1 g/mol+4×16.0 g/mol) = 0.025 mol×233.4 g/mol = 5.835 g
Therefore, 5.835 grams of barium sulfate will precipitate out in this reaction.
Question 17
Question
A solution contains 0.20 mol/L of silver nitrate (AgNO3) and 0.15 mol/L of
sodium chloride (NaCl). If these two solutions are mixed together, what is
the maximum concentration of silver chloride (AgCl) that could precipitate
out? Assume that the reaction goes to completion and that AgCl is the only
precipitate formed.
12
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by comparing the number of moles
of each reactant.
For AgNO3: Number of moles of AgNO3= 0.20 mol/L ×Vtotal volume Since
the stoichiometry is 1:1, the number of moles of AgNO3is equal to 0.20 ×
Vtotal volume mol.
For NaCl: Number of moles of NaCl = 0.15 mol/L ×Vtotal volume Since the
stoichiometry is 1:1, the number of moles of NaCl is equal to 0.15×Vtotal volume mol.
The limiting reactant will be the reactant which produces less AgCl.
Step 3: Calculate the maximum concentration of AgCl that could precipitate
out based on the limiting reactant.
Since the stoichiometry is 1:1 for all reactants and products, the number of
moles of AgCl that could form is equal to the number of moles of the limiting
reactant.
The maximum concentration of AgCl that could precipitate out would be
when both reactants are completely used up. So, the maximum concentration
of AgCl would be:
Max [AgCl] = minimum(0.15 ×Vtotal volume,0.20 ×Vtotal volume)
Therefore, the maximum concentration of AgCl that could precipitate out
depends on the total volume of the two solutions added.
Question 18
Question
Calculate the concentration of chloride ions (Cl−) in a solution formed by mixing
250 mL of 0.5 M NaCl and 500 mL of 0.25 M NaCl.
Solution
Step 1: Calculate the total moles of NaCl in each solution. Step 2: Determine
the total moles of NaCl after mixing the solutions. Step 3: Calculate the final
concentration of chloride ions in the mixed solution.
Step 1: For the first solution: Volume of NaCl solution = 250 mL = 0.25
L Molarity of NaCl solution = 0.5 M
Number of moles of NaCl in the first solution:
0.5 mol/L ×0.25 L = 0.125 mol
13
For the second solution: Volume of NaCl solution = 500 mL = 0.5 L Molarity
of NaCl solution = 0.25 M
Number of moles of NaCl in the second solution:
0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Total moles of NaCl after mixing the solutions:
0.125 mol + 0.125 mol = 0.25 mol
Step 3: Total volume of the mixed solution:
250 mL + 500 mL = 750 mL = 0.75 L
Concentration of chloride ions (Cl−) in the mixed solution:
0.25 mol
0.75 L = 0.33 M
Therefore, the concentration of chloride ions in the mixed solution is 0.33
M.
Question 19
Question
A solution is prepared by mixing 150 mL of 0.2 M lead(II) nitrate with 50 mL
of 0.3 M sodium chloride. a) Determine the balanced chemical equation for
the precipitation reaction that occurs when the two solutions are mixed. b)
Calculate the mass of lead(II) chloride precipitate that forms.
(Note: The molar mass of lead(II) chloride is 278.1 g/mol.)
Solution
a) First, let’s determine the balanced chemical equation for the precipitation
reaction between lead(II) nitrate and sodium chloride:
Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3
b) To find the mass of lead(II) chloride precipitate formed, we need to first
determine the limiting reactant. Let’s calculate the moles of each reactant:
Step 1: Calculate the moles of lead(II) nitrate: Moles of Pb(NO3)2= concen-
tration ×volume Moles of Pb(NO3)2= 0.2 mol/L ×0.15 L Moles of Pb(NO3)2
= 0.03 mol
Step 2: Calculate the moles of sodium chloride: Moles of NaCl = concen-
tration ×volume Moles of NaCl = 0.3 mol/L ×0.05 L Moles of NaCl = 0.015
mol
Step 3: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since the stoichiometry of the reaction is 1:2 for lead(II) nitrate
14
to lead(II) chloride, we need twice as many moles of lead(II) nitrate. Therefore,
sodium chloride is the limiting reactant.
Step 4: Calculate the mass of lead(II) chloride formed using the moles of
sodium chloride: Moles of PbCl2= Moles of NaCl (from Step 2) Moles of PbCl2
= 0.015 mol
Step 5: Calculate the mass of lead(II) chloride: Mass of PbCl2= moles of
PbCl2×molar mass of PbCl2Mass of PbCl2= 0.015 mol ×278.1 g/mol Mass
of PbCl2= 4.17 g
Therefore, the mass of lead(II) chloride precipitate that forms is 4.17 g.
Question 20
Question
Calculate the mass of lead(II) chloride (PbCl2) that can be produced from the
reaction between 50.0 mL of a 0.200 M lead(II) nitrate (Pb(NO3)2) solution and
excess hydrochloric acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and hydrochloric acid.
Pb(NO3)2(aq) + 2HCl(aq)→PbCl2(s) + 2HNO3(aq)
Step 2: Calculate the moles of lead(II) nitrate present in the solution using
the formula n= M ×V.
n(Pb(NO3)2)=M×V=0.200 M ×0.0500 L = 0.0100 mol
Step 3: Use the mole ratio from the balanced equation to find the moles of
lead chloride formed. Since the mole ratio is 1:1, the moles of lead chloride is
also 0.0100 mol.
Step 4: Calculate the mass of lead chloride formed using the molar mass of
PbCl2(Pb = 207.2 g/mol, Cl = 35.5 g/mol).
Molar mass of PbCl2= 207.2 + 2(35.5) = 278.2 g/mol
Mass of PbCl2=n×molar mass = 0.0100 mol ×278.2 g/mol = 2.78 g
Therefore, the mass of lead(II) chloride that can be produced is 2.78 grams.
15
Question 21
Question
A solution is prepared by mixing 100 mL of 0.2 M calcium chloride (CaCl2)
with 150 mL of 0.3 M sodium sulfate (Na2SO4). Determine the concentrations
of Ca2+ and SO2−
4ions in the resulting solution. (Assume complete dissociation
of the salts.)
Solution
Step 1: Calculate the moles of CaCl2. Given concentration of CaCl2= 0.2 M,
volume of CaCl2= 100 mL = 0.1 L Number of moles of CaCl2= concentration×
volume = 0.2 mol/L ×0.1 L = 0.02 mol
Step 2: Calculate the moles of Na2SO4. Given concentration of Na2SO4=
0.3 M, volume of Na2SO4= 150 mL = 0.15 L Number of moles of Na2SO4=
concentration ×volume = 0.3 mol/L ×0.15 L = 0.045 mol
Step 3: Determine the limiting reactant. The stoichiometry of the reaction
between CaCl2and Na2SO4is 1:1. Since the number of moles of CaCl2is less
than Na2SO4, CaCl2is the limiting reactant.
Step 4: Calculate the moles of products formed. For CaCl2, 1 mole yields 1
mole of Ca2+. Therefore, the moles of Ca2+ = 0.02 mol.
For Na2SO4, 1 mole yields 1 mole of SO2−
4. Therefore, the moles of SO2−
4=
0.02 mol.
Step 5: Calculate the concentrations of Ca2+ and SO2−
4ions in the resulting
solution. Total volume of the resulting solution = 100 mL + 150 mL = 250 mL
= 0.25 L
Concentration of Ca2+ =moles of Ca2+
total volume of solution =0.02 mol
0.25 L = 0.08 M
Concentration of SO2−
4=moles of SO2−
4
total volume of solution =0.02 mol
0.25 L = 0.08 M
Therefore, the concentrations of Ca2+ and SO2−
4ions in the resulting solution
are 0.08 M each.
Question 22
Question
A university laboratory needs to prepare a 250 mL solution that contains 0.10
M of sodium chloride (NaCl) and 0.20 M of silver nitrate (AgN O3). If these
two solutions are mixed together, how many grams of silver chloride (AgCl) will
precipitate out of the solution? Assume that the reaction goes to completion.
16
Solution
Step 1: Write the balanced chemical equation for the reaction between sodium
chloride and silver nitrate to form silver chloride:
NaCl(aq)+AgN O3(aq)→AgCl(s)+N aN O3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant initially. Given: Volume of NaCl solution = 250 mL = 0.250 L Molarity
of NaCl (M1) = 0.10 M Moles of N aCl:n1=M1×V= 0.10 mol/L×0.250 L=
0.025 mol
Molarity of AgNO3(M2) = 0.20 M Volume of AgN O3solution = 250 mL
= 0.250 L Moles of AgNO3:n2=M2×V= 0.20 mol/L ×0.250 L= 0.050 mol
Step 3: Determine the limiting reactant and calculate the moles of AgCl
formed. The balanced equation shows a 1:1 mole ratio between NaCl and AgCl.
Since the molar ratio between NaCl and AgCl is 1:1, the limiting reactant is
NaCl. Moles of AgCl formed: nAgCl =nN aCl = 0.025 mol
Step 4: Calculate the mass of AgCl formed using the molar mass of AgCl.
The molar mass of AgCl is calculated as the sum of the atomic masses of
silver (Ag) and chlorine (Cl): Ag = 107.87 g/mol,Cl = 35.45 g/mol AgCl =
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed: mAgCl =nAgCl×MAgCl = 0.025 mol×143.32 g/mol =
3.58 g
Therefore, 3.58 grams of silver chloride will precipitate out of the solution.
Question 23
Question
A solution is prepared by mixing 120 mL of 0.3 M silver nitrate (AgNO3) with
80 mL of 0.2 M sodium chloride (NaCl). Calculate the mass of silver chloride
(AgCl) precipitate that forms in grams. The reaction is given by the equation:
AgNO3+ NaCl →AgCl + NaNO3
Solution
Step 1: Determine the limiting reagent by comparing the number of moles of
each reactant. Let’s start by calculating the number of moles for each reactant.
The number of moles can be calculated using the formula:
Number of moles = Molarity ×Volume (in L)
For silver nitrate (AgNO3): Number of moles = 0.3 mol/L ×0.12 L = 0.036
mol
For sodium chloride (NaCl): Number of moles = 0.2 mol/L ×0.08 L = 0.016
mol
17
Step 2: Determine the limiting reagent. Since the reaction stoichiometry is
1:1 between silver nitrate and sodium chloride, the limiting reagent is the one
that produces the fewer moles of AgCl. In this case, sodium chloride is the
limiting reagent because it produces only 0.016 mol of AgCl compared to the
0.036 mol produced by silver nitrate.
Step 3: Calculate the mass of silver chloride formed. The molar mass of
AgCl is:
1×Ag + 1 ×Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Using the number of moles of AgCl formed (0.016 mol) and its molar mass,
we can calculate the mass of AgCl formed:
0.016 mol ×143.32 g/mol = 2.293 g
Therefore, the mass of silver chloride precipitate formed is 2.293 grams.
Question 24
Question
Calculate the mass of lead(II) iodide (PbI2) that can be produced when 50.0
mL of a 0.200 M lead(II) nitrate solution is mixed with 75.0 mL of a 0.150 M
sodium iodide solution. Assume that the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and sodium iodide to form lead(II) iodide and sodium nitrate.
Pb(NO3)2(aq) + 2NaI(aq)→PbI2(s) + 2NaNO3(aq)
Step 2: Calculate the amount of lead(II) nitrate and sodium iodide used.
Amount of Pb(NO3)2= Volume ×Molarity
= 0.0500 L ×0.200 mol/L
= 0.0100 mol
Amount of NaI = Volume ×Molarity
= 0.0750 L ×0.150 mol/L
= 0.0113 mol
Step 3: Determine the limiting reactant and calculate the maximum amount
of lead(II) iodide that can be produced. Since lead(II) nitrate and sodium iodide
react in a 1:2 molar ratio, the limiting reactant is the lead(II) nitrate.
Moles of PbI2= 0.0100 mol Pb(NO3)2×1 mol PbI2
1 mol Pb(NO3)2
= 0.0100 mol
18
Step 4: Convert moles of lead(II) iodide to mass.
Mass of PbI2= Moles ×Molar Mass
= 0.0100 mol ×(207.2 g/mol)
= 2.07 g
Therefore, the mass of lead(II) iodide that can be produced is 2.07 grams.
Question 25
Question
Calculate the mass of lead(II) iodide (PbI2) that can be formed when 50.0 mL
of 0.100 M lead(II) nitrate (Pb(NO3)2) solution is mixed with 50.0 mL of 0.100
M potassium iodide (KI) solution. Assume the reaction goes to completion and
that lead(II) iodide is the only product.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide.
Pb(NO3)2+ 2KI →PbI2+ 2KNO3
Step 2: Determine the limiting reactant by calculating the moles of each
reactant. For lead(II) nitrate:
moles = Volume ×Molarity
= (0.0500 L) ×(0.100 mol/L)
= 0.00500 mol
For potassium iodide:
moles = Volume ×Molarity
= (0.0500 L) ×(0.100 mol/L)
= 0.00500 mol
Both reactants have the same number of moles, so lead(II) nitrate is the
limiting reactant.
Step 3: Calculate the theoretical yield of lead(II) iodide by using the stoi-
chiometry from the balanced equation. From the balanced equation, 1 mole of
lead(II) nitrate produces 1 mole of lead(II) iodide.
moles of PbI2= moles of Pb(NO3)2= 0.00500 mol
19
Step 4: Convert moles of lead(II) iodide to mass using its molar mass.
Molar mass of PbI2= mass of Pb + 2 ×(mass of I)
= 207.2 g/mol + 2 ×126.9 g/mol
= 460.2 g/mol
Mass of PbI2= moles of PbI2×molar mass of PbI2
= 0.00500 mol ×460.2 g/mol
= 2.30 g
Therefore, the mass of lead(II) iodide that can be formed is 2.30 g.
Question 26
Question
Calculate the concentration of chloride ions in a solution that results from mix-
ing 100.0 mL of 0.200 M NaCl with 200.0 mL of 0.500 M MgCl2. Assume that
the volumes are additive.
Solution
Step 1: Find the moles of Cl−ions from NaCl. Given: Volume of NaCl solution,
V1= 100.0 mL = 0.100 L Concentration of NaCl, C1= 0.200 M
The moles of Cl−ions from NaCl can be calculated using the formula:
moles of ions = volume ×concentration
Therefore,
moles of Cl−from NaCl = 0.100 L ×0.200 M = 0.020 moles
Step 2: Find the moles of Cl−ions from MgCl2. Given: Volume of MgCl2
solution, V2= 200.0 mL = 0.200 L Concentration of MgCl2,C2= 0.500 M
The moles of Cl−ions from MgCl2can be calculated using the same formula:
moles of ions = volume ×concentration
Therefore,
moles of Cl−from MgCl2= 0.200 L ×0.500 M = 0.100 moles
Step 3: Find the total moles of Cl−ions. Since the volumes of the solutions
are additive, the total moles of Cl−ions is:
Total moles of Cl−= moles of Cl−from NaCl+moles of Cl−from MgCl2= 0.020 moles+0.100 moles = 0.120 moles
20
Step 4: Calculate the concentration of chloride ions in the final solution. The
total volume of the final solution is the sum of the volumes of the two solutions:
Vtotal = 0.100 L + 0.200 L = 0.300 L
Therefore, the concentration of chloride ions in the final solution is:
Total moles of Cl−
Vtotal
=0.120 moles
0.300 L = 0.40 M
Question 27
Question
A solution is prepared by mixing 50.0 mL of 0.200 M silver nitrate (AgNO3)
with 50.0 mL of 0.100 M sodium chloride (NaCl). Determine the mass of silver
chloride (AgCl) that will precipitate.
Solution
Step 1: Write the balanced chemical equation for the reaction of silver nitrate
(AgNO3) with sodium chloride (NaCl) to form silver chloride (AgCl) precipitate:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant to find the maximum amount of
precipitate that can form. Calculating the moles of each reactant:
Moles of AgNO3= Volume (L) ×Molarity = 0.050 L ×0.200 mol/L =
0.010 mol
Moles of NaCl = Volume (L) ×Molarity = 0.050 L×0.100 mol/L = 0.005 mol
From the balanced chemical equation, the mole ratio of AgNO3to NaCl is
1:1. Since the moles of NaCl are less, it is the limiting reactant.
Step 3: Calculate the mass of silver chloride (AgCl) precipitate formed using
the moles of limiting reactant:
Moles of NaCl = 0.005 mol
Molar mass of AgCl: 1 atom of Ag + 1 atom of Cl = 107.87 g/mol +
35.45 g/mol = 143.32 g/mol
Mass of AgCl = Moles ×Molar mass = 0.005 mol ×143.32 g/mol = 0.717 g
Therefore, the mass of silver chloride (AgCl) that will precipitate is 0.717
grams.
Question 28
Question
Calculate the concentration of chloride ions in a solution prepared by mixing
100.0 mL of a 0.200 M solution of calcium chloride with 200.0 mL of a 0.500 M
solution of sodium chloride.
21
Solution
Step 1: Write the balanced chemical equation for the dissociation of calcium
chloride and sodium chloride in water.
CaCl2→Ca2+ + 2Cl−
NaCl →Na++ Cl−
Step 2: Calculate the moles of chloride ions from each solution. For calcium
chloride:
Moles of Cl−= Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.020 mol
For sodium chloride:
Moles of Cl−= Volume ×Molarity = 0.200 L ×0.500 mol/L = 0.100 mol
Step 3: Calculate the total moles of chloride ions in the final solution.
Total moles of Cl−= 0.020 mol + 0.100 mol = 0.120 mol
Step 4: Calculate the final volume of the solution.
Final volume = 100.0 mL + 200.0 mL = 300.0 mL = 0.300 L
Step 5: Calculate the concentration of chloride ions in the final solution.
Concentration of Cl−=Total moles of Cl−
Final volume =0.120 mol
0.300 L = 0.400 M
Therefore, the concentration of chloride ions in the final solution is 0.400 M.
Question 29
Question
A chemist needs to prepare 250 mL of a 0.05 M solution of silver nitrate
(AgNO3) by dissolving solid AgN O3in distilled water. The only source avail-
able is AgNO3solid and the chemist estimates the molar mass of AgNO3to
be 170.9 g/mol. What mass of AgNO3should the chemist dissolve in water to
prepare the desired solution?
Solution
Step 1: Calculate the number of moles of AgN O3required to make 250 mL of
a 0.05 M solution. Step 2: Use the molar mass of AgNO3to find the mass of
AgNO3required to prepare the solution.
Step 1: Given: Volume of solution to be prepared, V= 250 mL = 0.250 L
Molarity of the solution, M= 0.05 mol/L
22
Using the definition of molarity, which is given by M=n
V, where nis the
number of moles and Vis the volume in liters, we can rearrange the equation
to solve for n:
n=M×V n = 0.05 mol/L ×0.250 L n= 0.0125 mol
Therefore, the chemist needs 0.0125 moles of AgN O3to prepare the solution.
Step 2: Given: Molar mass of AgNO3,MM = 170.9 g/mol
To find the mass of AgNO3required, we can use the formula:
Mass = Number of moles ×Molar mass
Substitute the known values:
Mass = 0.0125 mol ×170.9 g/mol Mass = 2.13625 g
Therefore, the chemist should dissolve 2.13625 g of AgN O3in water to
prepare the desired 0.05 M solution.
Question 30
Question
A student is conducting an experiment to determine the concentration of chlo-
ride ions in a water sample. The student adds excess silver nitrate solution to
100.0 mL of the water sample, which precipitates all the chloride ions as sil-
ver chloride. After filtration, the mass of the silver chloride obtained is 0.563
g. Calculate the concentration of chloride ions in the original water sample in
units of mol/L.
(Silver chloride has a molar mass of 143.32 g/mol)
Solution
Step 1: Calculate the moles of silver chloride formed.
Moles of silver chloride = Mass of silver chloride
Molar mass =0.563 g
143.32 g/mol
Step 2: Calculate the moles of chloride ions in the silver chloride. 1 mole of
silver chloride contains 1 mole of chloride ions.
Moles of chloride ions = Moles of silver chloride
Step 3: Calculate the volume of the water sample in liters. Volume of water
sample = 100.0 mL = 0.100 L
Step 4: Calculate the concentration of chloride ions in the original water
sample.
Concentration of chloride ions = Moles of chloride ions
Volume of water sample
23
Question 31
Question
Calculate the concentration of chloride ions in a solution when 100.0 mL of 0.200
M silver nitrate (AgNO3) is mixed with 100.0 mL of 0.150 M sodium chloride
(NaCl). The Ksp of silver chloride (AgCl) is 1.8×10−10.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant. Since both reactants have a 1:1
molar ratio in the balanced chemical equation, the reactant that produces less
silver chloride will be the limiting reactant.
Calculate the moles of AgNO3:
Moles of AgNO3= Volume ×Molarity
Moles of AgNO3= 0.100 L ×0.200 mol/L = 0.020 mol
Calculate the moles of NaCl:
Moles of NaCl = Volume ×Molarity
Moles of NaCl = 0.100 L ×0.150 mol/L = 0.015 mol
Since 0.015 moles of NaCl produces less silver chloride than 0.020 moles of
AgNO3, NaCl is the limiting reactant.
Step 3: Determine the moles of AgCl formed. From the balanced equation,
1 mole of NaCl forms 1 mole of AgCl. So, 0.015 moles of NaCl will form 0.015
moles of AgCl.
Step 4: Calculate the concentration of Cl−ions in the solution. The total
volume of the solution is 200.0 mL or 0.200 L. Therefore, the concentration of
Cl−ions in the solution is:
Concentration of Cl−=Moles of AgCl
Total Volume
Concentration of Cl−=0.015 mol
0.200 L = 0.075 mol/L
The concentration of chloride ions in the solution is 0.075 mol/L.
Question 32
Question
A solution contains 0.2 M of calcium chloride. If we add 0.1 M of sodium sulfate
to the solution, what is the concentration of calcium ions after precipitation
occurs? The Ksp of calcium sulfate is 2.4×10−5.
24
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between calcium chloride and sodium sulfate:
CaCl2(aq) + Na2SO4(aq)→CaSO4(s) + 2NaCl(aq)
Step 2: Calculate the initial concentration of calcium ions, Ca2+, beforetheprecipitationreactionoccurs :
[Ca2+]initial = 0.2 M
Step 3: Determine the concentration of calcium ions, Ca2+, aftertheprecipitationreactionoccurs.LetxbetheconcentrationofCa2+ionsatequilibrium.T hebalancedchemicalequationshowsa1 :
1molarratiobetweenCaCl2andCaSO4.T heref ore, theconcentrationof Ca2+ionsafterprecipitationis :
[Ca2+]final = 0.2 M + x
Step 4: Write the expression for the solubility product,
Ksp = [Ca2+][SO2−
4]
, and substitute the known values:
2.4×10−5= (0.2 + x)(0.1)
Step 5: Solve for x to find the concentration of calcium ions after precipita-
tion occurs:
2.4×10−5= 0.02 + 0.1x
0.1x= 2.4×10−5−0.02
x≈0.00016 M
Therefore, the concentration of calcium ions after precipitation occurs is
approximately 0.00016 M.
Question 33
Question
During a rainstorm, 2 inches of rain falls over a 1-acre area. If 1 inch of rain
is equivalent to 4.168 cubic feet of water, how many gallons of water fell on
the 1-acre area during the rainstorm? (Hint: 1 gallon is equivalent to 0.133681
cubic feet)
Solution
Step 1: Find the total volume of water that fell on the 1-acre area in cubic feet.
Since 1 inch of rain is equivalent to 4.168 cubic feet of water, the total volume
of water that fell on the 1-acre area can be calculated as follows:
2 inches ×4.168 cubic feet per inch = 8.336 cubic feet
25
Step 2: Convert the total volume of water to gallons. Since 1 gallon is
equivalent to 0.133681 cubic feet, we can convert the total volume of water in
cubic feet to gallons as follows:
8.336 cubic feet
0.133681 cubic feet per gallon ≈62.37 gallons
Therefore, approximately 62.37 gallons of water fell on the 1-acre area during
the rainstorm.
Question 34
Question
A chemistry student is performing a precipitation reaction between aqueous
solutions of lead(II) nitrate and potassium iodide. If 50.0 mL of 0.200 M lead(II)
nitrate is mixed with excess potassium iodide, what mass of lead(II) iodide will
precipitate?
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and potassium iodide. The balanced equation is:
Pb(NO3)2(aq) + 2KI(aq) →PbI2(s) + 2KNO3(aq)
Step 2: Determine the limiting reactant by calculating the amount of lead(II)
iodide that would be formed from each reactant. Since lead(II) nitrate and
potassium iodide react in a 1:2 mole ratio, the moles of lead(II) nitrate is given
by:
moles = Molarity ×Volume = 0.200 mol/L ×0.0500 L = 0.0100 mol
Similarly, the moles of lead(II) iodide formed by lead(II) nitrate is:
moles PbI2= 0.0100 mol
Step 3: Calculate the mass of lead(II) iodide formed using the molar mass
of lead(II) iodide (461 g/mol):
Mass = mol ×Molar mass = 0.0100 mol ×461 g/mol = 4.61 g
Therefore, the mass of lead(II) iodide that will precipitate is 4.61 grams.
26
Question 35
Question
A chemistry student is conducting an experiment where a silver nitrate solution
reacts with a sodium chloride solution to produce silver chloride precipitate.
The student mixes 100.0 mL of 0.200 M silver nitrate solution with 150.0 mL of
0.150 M sodium chloride solution. What mass of silver chloride precipitate will
form?
(Hint: Silver chloride is insoluble in water and will precipitate out of solu-
tion).
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3+NaCl →AgCl +N aN O3
Step 2: Determine the limiting reactant by comparing the number of moles
of each reactant. The number of moles of silver nitrate:
moles of AgNO3= Molarity ×Volume (L)
moles of AgNO3= 0.200 mol/L ×0.100 L = 0.0200 mol
The number of moles of sodium chloride:
moles of NaCl = Molarity ×Volume (L)
moles of NaCl = 0.150 mol/L ×0.150 L = 0.0225 mol
Since silver nitrate and sodium chloride react in the ratio of 1:1, the limiting
reactant is silver nitrate because it produces fewer moles of product.
Step 3: Calculate the mass of silver chloride precipitate that will form. The
molar mass of silver chloride (AgCl) is approximately 143.32 g/mol.
moles of AgCl = moles of AgNO3
mass of AgCl = moles of AgCl ×molar mass of AgCl
mass of AgCl = 0.0200 mol ×143.32 g/mol = 2.8664 g
Therefore, the mass of silver chloride precipitate that will form is 2.8664
grams.
27
Question 2
Question
A solution contains 250 mL of 0.2 M silver nitrate (AgNO3) and 300 mL of 0.3
M sodium chloride (NaCl). Calculate the mass of silver chloride (AgCl) that
will precipitate when the two solutions are mixed.
(Hint: AgCl precipitates according to the following reaction: AgNO3+
NaCl →AgCl ↓+NaNO3)
Solution
Step 1: Calculate the moles of AgNO3and NaCl using the formula n= M ×V.
For AgNO3:
nAgNO3= 0.2 mol/L ×0.250 L = 0.05 mol
For NaCl:
nNaCl = 0.3 mol/L ×0.300 L = 0.09 mol
Step 2: Determine the limiting reactant by comparing the moles of each
reactant. According to the balanced chemical equation, 1 mole of AgNO3reacts
with 1 mole of NaCl to form 1 mole of AgCl. Since the mole ratio is 1:1, the
limiting reactant is the one that produces the least amount of product. In this
case, AgNO3is the limiting reactant because it produces 0.05 moles of AgCl
compared to 0.09 moles from NaCl.
Step 3: Calculate the mass of AgCl precipitated using the molar mass of
AgCl (143.32 g/mol).
Mass of AgCl = nAgCl×Molar mass of AgCl = 0.05 mol×143.32 g/mol = 7.166 g
Therefore, when 250 mL of 0.2 M AgNO3is added to 300 mL of 0.3 M NaCl,
7.166 g of AgCl will precipitate.
Question 3
Question
Calculate the solubility of lead(II) iodide (PbI2) at 25
°
C. The Ksp of PbI2is
7.1×10−9.
Solution
Step 1: Write the solubility equilibrium expression for lead(II) iodide:
PbI2⇌Pb2+ + 2I−
The equilibrium constant expression for this reaction is:
Ksp = [Pb2+][I−]2
2
Step 2: Let’s assume that the solubility of PbI2is x. This means the equi-
librium concentrations of Pb2+ and I−will be xand 2xrespectively.
Step 3: Substitute the equilibrium concentrations into the Ksp expression:
7.1×10−9= (x)(2x)2
Step 4: Solve for x:
7.1×10−9= 4x3
x3=7.1×10−9
4
x=3
r7.1×10−9
4
Step 5: Calculate the solubility of lead(II) iodide:
x≈5.72 ×10−3M
Therefore, the solubility of lead(II) iodide at 25
°
C is approximately 5.72 ×
10−3M.
Question 4
Question
Calculate the molarity of barium chloride (BaCl2) solution when 50.0 mL of a
0.200 M solution of silver nitrate (AgNO3) is added to 75.0 mL of a solution
containing 0.150 M of sodium sulfate (Na2SO4). Assume the reaction goes to
completion and that barium sulfate (BaSO4) precipitates.
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium sulfate to find the net ionic equation.
AgNO3+ Na2SO4→Ag2SO4+ NaNO3
The net ionic equation is:
Ag++ SO2−
4→Ag2SO4
Step 2: Determine the limiting reagent to find the moles of barium sulfate
formed.
From the net ionic equation, it is clear that 1 mole of barium sulfate is
formed from 1 mole of silver nitrate.
Moles of silver nitrate = 0.0500 L ×0.200 mol/L = 0.0100 mol
Since the reaction ratio is 1:1, moles of barium sulfate formed = 0.0100 mol
Step 3: Calculate the moles of barium chloride formed using the mole ratio
from the balanced equation.
From the balanced equation for the precipitation reaction:
BaCl2+ Na2SO4→BaSO4+ NaCl
3
We have a 1:1 mole ratio between barium chloride and barium sulfate.
Therefore, moles of barium chloride = moles of barium sulfate = 0.0100 mol
Step 4: Calculate the molarity of the barium chloride solution.
Volume of barium chloride solution = 75.0 mL + 50.0 mL = 125.0 mL =
0.1250 L
Molarity of barium chloride solution = 0.0100 mol
0.1250 L = 0.0800 M
Therefore, the molarity of the barium chloride solution is 0.0800 M.
Question 5
Question
In a lab experiment, a solution of 50 mL of 0.2 M potassium iodide (KI) is mixed
with 30 mL of 0.3 M lead(II) nitrate (Pb(NO3)2) solution. The only product
formed is lead(II) iodide (PbI2), which precipitates out of the solution.
Calculate the maximum mass of lead(II) iodide that can be formed in grams.
(Hint: The balanced chemical equation for the reaction is Pb(NO3)2+2KI →
PbI2+ 2KNO3)
Solution
Step 1: Calculate the moles of potassium iodide (KI) and lead(II) nitrate
(Pb(NO3)2) used in the reaction. Given: For potassium iodide (KI): Volume =
50 mL = 0.05 L Molarity = 0.2 M
Number of moles of KI = Molarity ×Volume Number of moles of KI =
0.2 mol/L ×0.05 L = 0.01 mol
For lead(II) nitrate (Pb(NO3)2): Volume = 30 mL = 0.03 L Molarity = 0.3
M
Number of moles of Pb(NO3)2= Molarity ×Volume Number of moles of
Pb(NO3)2= 0.3 mol/L ×0.03 L = 0.009 mol
Step 2: Determine the limiting reactant. From the balanced chemical equa-
tion, 1 mole of Pb(NO3)2reacts with 2 moles of KI to form 1 mole of PbI2.
Since the ratio is 1:2, Pb(NO3)2is the limiting reactant because it produces
fewer moles of PbI2compared to KI.
Step 3: Calculate the maximum mass of lead(II) iodide (PbI2) that can be
formed. From the stoichiometry of the reaction, 1 mole of PbI2has a molar
mass of 461 g/mol. Number of moles of PbI2= 0.009 mol (since Pb(NO3)2is
the limiting reactant)
Maximum mass of PbI2= Number of moles ×Molar mass Maximum mass
of PbI2= 0.009 mol ×461 g/mol = 4.149 g
Therefore, the maximum mass of lead(II) iodide that can be formed is 4.149
grams.
4
Question 6
Question
Calculate the solubility of silver chloride (AgCl) in a 0.010 M solution of silver
nitrate (AgNO3). The Ksp of AgCl is 1.8×10−10.
Solution
Step 1: Write the balanced equation for the dissociation of AgCl:
AgCl ⇌Ag++ Cl−
Step 2: Set up an ICE (Initial, Change, Equilibrium) table for the dissocia-
tion of AgCl:
Substance AgCl Ag+Cl−
Initial (M) x0 0
Change (M) −x+x+x
Equilibrium (M) x−x x x
Step 3: Write the expression for the solubility product Ksp:
Ksp = [Ag+][Cl−]
Step 4: Substitute the equilibrium concentrations into the Ksp expression:
1.8×10−10 =x×x
Step 5: Solve for xto find the equilibrium concentration of Ag+and Cl−
ions:
x=p1.8×10−10 = 1.34 ×10−5
Step 6: The solubility of AgCl in the 0.010 M solution of AgNO3is equal to
the concentration of Ag+ions, which is 1.34 ×10−5M.
Question 7
Question
Calculate the mass of lead(II) iodide that will precipitate when 50.0 mL of 0.200
M lead(II) nitrate is mixed with 25.0 mL of 0.300 M potassium iodide. Assume
the reaction goes to completion.
Solution
Step 1: Write down the balanced chemical equation for the reaction.
Pb(NO3)2(aq) + 2KI (aq) →PbI2(s) + 2KNO3(aq)
5
Step 2: Determine the limiting reactant. Calculate the moles of lead(II)
nitrate and potassium iodide: For lead(II) nitrate: (0.200 M) ×(0.0500 L) =
0.0100 mol For potassium iodide: (0.300 M) ×(0.0250 L) = 0.00750 mol
Since potassium iodide is limiting (less moles), it will completely react.
Step 3: Calculate the moles of lead(II) iodide precipitated. Using stoichiom-
etry from the balanced equation: 1 mol of Pb(NO3)2produces 1 mol of PbI2
Therefore, 0.00750 mol of PbI2will be formed.
Step 4: Convert moles of lead(II) iodide to mass. Given molar mass of PbI2
is approximately 461.01 g/mol, Mass of PbI2= 0.00750 mol ×461.01 g/mol =
3.46 g
So, the mass of lead(II) iodide that will precipitate is 3.46 grams.
Question 8
Question
A researcher is studying the precipitation patterns in a region and collects the
following data for the month of April:
Total rainfall: 120 mm
Number of rainy days: 15
Assuming that the rainfall is evenly distributed over the rainy days, deter-
mine the average precipitation intensity in mm/day during the month of April.
Solution
To find the average precipitation intensity in mm/day during the month of April,
we need to divide the total rainfall by the number of rainy days.
Step 1: Calculate the average precipitation intensity.
Average precipitation intensity = Total rainfall
Number of rainy days
Average precipitation intensity = 120 mm
15 = 8 mm/day
Step 2: Answer: The average precipitation intensity during the month of
April is 8 mm/day.
Question 9
Question
A chemist needs to prepare 500 mL of a 0.2 M lead nitrate (Pb(NO3)2) solution
by dissolving solid lead nitrate in water. If the molar mass of lead nitrate is
331.21 g/mol, determine the mass of lead nitrate needed.
6
Solution
Step 1: Calculate the number of moles of lead nitrate needed. We can use the
formula M=n
V, where Mis the molarity, nis the number of moles, and Vis
the volume in liters.
Given: M= 0.2 M V= 500 mL = 0.5 L
Plugging in the values: n=M×V= 0.2 mol/L ×0.5 L = 0.1 mol
Therefore, the chemist needs 0.1 moles of lead nitrate.
Step 2: Calculate the mass of lead nitrate needed. We can use the formula
m=n×MM, where mis the mass, nis the number of moles, and MM is the
molar mass.
Given: n= 0.1 mol MM = 331.21 g/mol
Plugging in the values: m= 0.1 mol ×331.21 g/mol = 33.121 g
Therefore, the chemist needs 33.121 grams of lead nitrate to prepare 500 mL
of a 0.2 M solution.
Question 10
Question
A chemistry student wants to determine the concentration of chloride ions in a
solution. To do this, they add silver nitrate (AgNO3) to a 100.0 mL sample of
the solution, causing a white precipitate of silver chloride (AgCl) to form. The
student then filters the precipitate, dries it, and weighs it. The mass of the dry
silver chloride obtained is 0.345 g. Calculate the concentration of chloride ions
in the original solution in units of mol/L. The molar mass of AgCl is 143.32
g/mol.
Solution
Step 1: Write the balanced chemical equation for the reaction that occurs when
silver nitrate is added to chloride ions in solution. The balanced chemical equa-
tion is:
Ag++ Cl−→AgCl
Step 2: Calculate the moles of AgCl formed. Given: Mass of AgCl = 0.345
g Molar mass of AgCl = 143.32 g/mol
Moles of AgCl = Mass
Molar mass =0.345 g
143.32 g/mol ≈0.0024 mol
Step 3: Calculate the moles of chloride ions in the original solution. Since 1
mole of AgCl is produced when 1 mole of chloride ion is present, the moles of
chloride ions in the original solution is also 0.0024 mol.
Step 4: Determine the volume of the original solution. Given: Volume of
solution = 100.0 mL = 0.100 L
7
Step 5: Calculate the concentration of chloride ions in the original solution.
The concentration of chloride ions is given by:
Concentration = Moles of solute
Volume of solution (in L) =0.0024 mol
0.100 L = 0.024 mol/L
Answer: The concentration of chloride ions in the original solution is 0.024
mol/L.
Question 11
Question
Calculate the concentration of a new precipitate formed when 50.0 mL of a 0.200
M solution of calcium chloride is mixed with 100.0 mL of a 0.300 M solution of
sodium carbonate. Assume the reaction goes to completion and that the volume
of the final solution is 150.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride (CaCl2) and sodium carbonate (Na2CO3):
CaCl2+ Na2CO3→CaCO3+ 2NaCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant: For calcium chloride: Number of moles = Concentration ×Volume
= 0.200 M ×0.0500 L = 0.0100 mol For sodium carbonate: Number of moles =
Concentration ×Volume = 0.300 M ×0.100 L = 0.0300 mol
Since calcium chloride produces 1 mol of precipitate for every 1 mol of sodium
carbonate, calcium chloride is the limiting reactant.
Step 3: Calculate the number of moles of precipitate formed (calcium car-
bonate): Number of moles of CaCO3= Number of moles of limiting reactant
= 0.0100 mol
Step 4: Calculate the concentration of the precipitate in the final solution:
Total volume of final solution = 150.0 mL = 0.150 L Concentration of CaCO3
=0.0100 mol
0.150 L Concentration of precipitate = 0.067 M
Therefore, the concentration of the new precipitate formed in the final solu-
tion is 0.067 M.
Question 12
Question
Calculate the mass of silver chloride (AgCl) that will precipitate when a so-
lution containing 0.050 moles of silver nitrate (AgNO3) is added to a solution
containing 0.035 moles of sodium chloride (NaCl). Assume that the reaction
goes to completion.
8
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate and sodium chloride:
AgNO3+ NaCl →AgCl + NaNO3
Step 2: Determine the limiting reactant to find the maximum amount of
silver chloride that can be formed. First, calculate the theoretical yield of AgCl
using the moles of AgNO3:
moles of AgCl = moles of AgNO3= 0.050 mol
Step 3: Calculate the theoretical yield of AgCl using the moles of NaCl:
moles of AgCl = moles of NaCl = 0.035 mol
Step 4: Identify the limiting reactant by comparing the amounts of AgCl
produced from each reactant. Since AgNO3produces more AgCl (0.050 moles)
than NaCl (0.035 moles), NaCl is the limiting reactant.
Step 5: Calculate the mass of silver chloride precipitated from the limiting
reactant (sodium chloride):
molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
mass of AgCl = moles of AgCl ×molar mass of AgCl
mass of AgCl = 0.035 mol ×143.32 g/mol = 5.0132 g
Thus, the mass of silver chloride that will precipitate when 0.050 moles of
silver nitrate is added to a solution containing 0.035 moles of sodium chloride
is 5.0132 grams.
Question 13
Question
A solution is prepared by mixing 100 mL of a 0.2 M calcium chloride solution
with 200 mL of a 0.4 M sodium sulfate solution. Determine if a precipitate will
form when these two solutions are mixed.
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride (CaCl2) and sodium sulfate (Na2SO4).
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Calculate the initial moles of each ion in the solutions. For calcium
chloride solution: - Moles of Ca2+ ions: 0.2 M ×0.1 L = 0.02 moles - Moles of
Cl−ions: 0.2 M ×0.1 L = 0.02 moles
9
For sodium sulfate solution: - Moles of Na+ions: 0.4 M ×0.2 L = 0.08 moles
- Moles of SO2−
4ions: 0.4 M ×0.2 L = 0.08 moles
Step 3: Determine the limiting reactant by examining the mole ratio in the
balanced equation. From the equation, the mole ratio of CaCl2to Na2SO4is
1:1. Therefore, both reactants are in stoichiometric amounts and there is no
limiting reactant.
Step 4: Write the net ionic equation for the precipitation reaction.
Ca2+ + SO2−
4→CaSO4
Step 5: Check the solubility rules to determine if CaSO4is insoluble. Cal-
cium sulfate (CaSO4) is insoluble according to most solubility rules, and thus a
precipitate will form when the two solutions are mixed.
Therefore, when the 0.2 M calcium chloride solution is mixed with the 0.4
M sodium sulfate solution, a white precipitate of calcium sulfate will form.
Question 14
Question
Calculate the mass of lead(II) chloride (P bCl2) that can be formed when 500.0
mL of 0.200 M lead(II) nitrate (P b(NO3)2) solution is reacted with excess
sodium chloride (NaCl). Assume the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and sodium chloride to form lead(II) chloride.
P b(NO3)2+ 2NaCl →P bCl2+ 2NaNO3
Step 2: Calculate the moles of lead(II) nitrate using the given volume and
molarity.
moles = volume ×molarity = 0.5 L ×0.200 M = 0.100 moles
Step 3: Use the mole ratio from the balanced equation to find the moles of
lead(II) chloride that can be formed.
moles of PbCl2= 0.100 moles ×1 mol PbCl2
1 mol Pb(NO3)2
= 0.100 moles
Step 4: Calculate the mass of lead(II) chloride formed using the molar mass
of P bCl2.
molar mass of PbCl2= 207.2 g/mol + 2 ×35.45 g/mol = 278.1 g/mol
mass of PbCl2= 0.100 moles ×278.1 g/mol = 27.81 g
Answer: The mass of lead(II) chloride that can be formed is 27.81 grams.
10
Question 15
Question
A solution contains 100 mL of 0.2 M calcium chloride and 200 mL of 0.15 M
sodium sulfate. What mass of calcium sulfate can be precipitated out?
(Hint: Use stoichiometry to calculate the limiting reactant and then deter-
mine the mass of the precipitate.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between calcium chloride (CaCl2) and sodium sulfate (Na2SO4) to form calcium
sulfate (CaSO4) and sodium chloride (NaCl).
CaCl2+ Na2SO4→CaSO4+ 2NaCl
Step 2: Calculate the moles of each reactant using the molarity formula
M=moles
liters . For calcium chloride:
0.2 M = molesCaCl2
0.1 L
molesCaCl2= 0.2 mol
For sodium sulfate:
0.15 M = molesNa2SO4
0.2 L
molesNa2SO4= 0.03 mol
Step 3: Determine the limiting reactant by comparing the moles of each
reactant and their stoichiometric coefficients. Since 1 mole of calcium chloride
reacts with 1 mole of sodium sulfate, whichever reactant has fewer moles is the
limiting reactant. In this case, sodium sulfate is the limiting reactant.
Step 4: Calculate the moles of calcium sulfate that can be formed based on
the limiting reactant (sodium sulfate). From the reaction, 1 mole of sodium
sulfate reacts to form 1 mole of calcium sulfate.
molesCaSO4= 0.03 mol
Step 5: Calculate the mass of calcium sulfate precipitated using the molar
mass of calcium sulfate (CaSO4) which is 136.14 g/mol.
MassCaSO4= molesCaSO4×Molar massCaSO4
MassCaSO4= 0.03 mol ×136.14 g/mol = 4.0842 g
Therefore, approximately 4.08 g of calcium sulfate can be precipitated out.
11
Question 16
Question
A university chemistry lab is conducting an experiment where a solution of
barium chloride (BaCl2) is mixed with a solution of sulfuric acid (H2SO4). If
100.0 mL of 0.200 M barium chloride solution is mixed with 50.0 mL of 0.500
M sulfuric acid solution, what mass of barium sulfate (BaSO4) will precipitate
out?
(Hint: The reaction is: BaCl2+ H2SO4→BaSO4+ 2HCl.)
Solution
Step 1: Write the balanced chemical equation for the reaction between barium
chloride and sulfuric acid.
BaCl2+ H2SO4→BaSO4+ 2HCl
Step 2: Calculate the moles of each reactant used in the reaction.
Moles of barium chloride:
Moles of BaCl2= Volume (L)×Molarity = 0.100 L×0.200 mol/L = 0.020 mol
Moles of sulfuric acid:
Moles of H2SO4= Volume (L)×Molarity = 0.050 L×0.500 mol/L = 0.025 mol
Step 3: Determine the limiting reactant. Since the reaction requires 1 mole
of BaCl2to react with 1 mole of H2SO4, and there are more moles of BaCl2
available, H2SO4is the limiting reactant.
Step 4: Calculate the mass of barium sulfate precipitated out using the
limiting reactant.
Moles of BaSO4= Moles of limiting reactant = 0.025 mol
Mass of BaSO4= Moles×Molar mass = 0.025 mol×(137.3 g/mol+32.1 g/mol+4×16.0 g/mol) = 0.025 mol×233.4 g/mol = 5.835 g
Therefore, 5.835 grams of barium sulfate will precipitate out in this reaction.
Question 17
Question
A solution contains 0.20 mol/L of silver nitrate (AgNO3) and 0.15 mol/L of
sodium chloride (NaCl). If these two solutions are mixed together, what is
the maximum concentration of silver chloride (AgCl) that could precipitate
out? Assume that the reaction goes to completion and that AgCl is the only
precipitate formed.
12
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by comparing the number of moles
of each reactant.
For AgNO3: Number of moles of AgNO3= 0.20 mol/L ×Vtotal volume Since
the stoichiometry is 1:1, the number of moles of AgNO3is equal to 0.20 ×
Vtotal volume mol.
For NaCl: Number of moles of NaCl = 0.15 mol/L ×Vtotal volume Since the
stoichiometry is 1:1, the number of moles of NaCl is equal to 0.15×Vtotal volume mol.
The limiting reactant will be the reactant which produces less AgCl.
Step 3: Calculate the maximum concentration of AgCl that could precipitate
out based on the limiting reactant.
Since the stoichiometry is 1:1 for all reactants and products, the number of
moles of AgCl that could form is equal to the number of moles of the limiting
reactant.
The maximum concentration of AgCl that could precipitate out would be
when both reactants are completely used up. So, the maximum concentration
of AgCl would be:
Max [AgCl] = minimum(0.15 ×Vtotal volume,0.20 ×Vtotal volume)
Therefore, the maximum concentration of AgCl that could precipitate out
depends on the total volume of the two solutions added.
Question 18
Question
Calculate the concentration of chloride ions (Cl−) in a solution formed by mixing
250 mL of 0.5 M NaCl and 500 mL of 0.25 M NaCl.
Solution
Step 1: Calculate the total moles of NaCl in each solution. Step 2: Determine
the total moles of NaCl after mixing the solutions. Step 3: Calculate the final
concentration of chloride ions in the mixed solution.
Step 1: For the first solution: Volume of NaCl solution = 250 mL = 0.25
L Molarity of NaCl solution = 0.5 M
Number of moles of NaCl in the first solution:
0.5 mol/L ×0.25 L = 0.125 mol
13
For the second solution: Volume of NaCl solution = 500 mL = 0.5 L Molarity
of NaCl solution = 0.25 M
Number of moles of NaCl in the second solution:
0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Total moles of NaCl after mixing the solutions:
0.125 mol + 0.125 mol = 0.25 mol
Step 3: Total volume of the mixed solution:
250 mL + 500 mL = 750 mL = 0.75 L
Concentration of chloride ions (Cl−) in the mixed solution:
0.25 mol
0.75 L = 0.33 M
Therefore, the concentration of chloride ions in the mixed solution is 0.33
M.
Question 19
Question
A solution is prepared by mixing 150 mL of 0.2 M lead(II) nitrate with 50 mL
of 0.3 M sodium chloride. a) Determine the balanced chemical equation for
the precipitation reaction that occurs when the two solutions are mixed. b)
Calculate the mass of lead(II) chloride precipitate that forms.
(Note: The molar mass of lead(II) chloride is 278.1 g/mol.)
Solution
a) First, let’s determine the balanced chemical equation for the precipitation
reaction between lead(II) nitrate and sodium chloride:
Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3
b) To find the mass of lead(II) chloride precipitate formed, we need to first
determine the limiting reactant. Let’s calculate the moles of each reactant:
Step 1: Calculate the moles of lead(II) nitrate: Moles of Pb(NO3)2= concen-
tration ×volume Moles of Pb(NO3)2= 0.2 mol/L ×0.15 L Moles of Pb(NO3)2
= 0.03 mol
Step 2: Calculate the moles of sodium chloride: Moles of NaCl = concen-
tration ×volume Moles of NaCl = 0.3 mol/L ×0.05 L Moles of NaCl = 0.015
mol
Step 3: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since the stoichiometry of the reaction is 1:2 for lead(II) nitrate
14
to lead(II) chloride, we need twice as many moles of lead(II) nitrate. Therefore,
sodium chloride is the limiting reactant.
Step 4: Calculate the mass of lead(II) chloride formed using the moles of
sodium chloride: Moles of PbCl2= Moles of NaCl (from Step 2) Moles of PbCl2
= 0.015 mol
Step 5: Calculate the mass of lead(II) chloride: Mass of PbCl2= moles of
PbCl2×molar mass of PbCl2Mass of PbCl2= 0.015 mol ×278.1 g/mol Mass
of PbCl2= 4.17 g
Therefore, the mass of lead(II) chloride precipitate that forms is 4.17 g.
Question 20
Question
Calculate the mass of lead(II) chloride (PbCl2) that can be produced from the
reaction between 50.0 mL of a 0.200 M lead(II) nitrate (Pb(NO3)2) solution and
excess hydrochloric acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and hydrochloric acid.
Pb(NO3)2(aq) + 2HCl(aq)→PbCl2(s) + 2HNO3(aq)
Step 2: Calculate the moles of lead(II) nitrate present in the solution using
the formula n= M ×V.
n(Pb(NO3)2)=M×V=0.200 M ×0.0500 L = 0.0100 mol
Step 3: Use the mole ratio from the balanced equation to find the moles of
lead chloride formed. Since the mole ratio is 1:1, the moles of lead chloride is
also 0.0100 mol.
Step 4: Calculate the mass of lead chloride formed using the molar mass of
PbCl2(Pb = 207.2 g/mol, Cl = 35.5 g/mol).
Molar mass of PbCl2= 207.2 + 2(35.5) = 278.2 g/mol
Mass of PbCl2=n×molar mass = 0.0100 mol ×278.2 g/mol = 2.78 g
Therefore, the mass of lead(II) chloride that can be produced is 2.78 grams.
15
Question 21
Question
A solution is prepared by mixing 100 mL of 0.2 M calcium chloride (CaCl2)
with 150 mL of 0.3 M sodium sulfate (Na2SO4). Determine the concentrations
of Ca2+ and SO2−
4ions in the resulting solution. (Assume complete dissociation
of the salts.)
Solution
Step 1: Calculate the moles of CaCl2. Given concentration of CaCl2= 0.2 M,
volume of CaCl2= 100 mL = 0.1 L Number of moles of CaCl2= concentration×
volume = 0.2 mol/L ×0.1 L = 0.02 mol
Step 2: Calculate the moles of Na2SO4. Given concentration of Na2SO4=
0.3 M, volume of Na2SO4= 150 mL = 0.15 L Number of moles of Na2SO4=
concentration ×volume = 0.3 mol/L ×0.15 L = 0.045 mol
Step 3: Determine the limiting reactant. The stoichiometry of the reaction
between CaCl2and Na2SO4is 1:1. Since the number of moles of CaCl2is less
than Na2SO4, CaCl2is the limiting reactant.
Step 4: Calculate the moles of products formed. For CaCl2, 1 mole yields 1
mole of Ca2+. Therefore, the moles of Ca2+ = 0.02 mol.
For Na2SO4, 1 mole yields 1 mole of SO2−
4. Therefore, the moles of SO2−
4=
0.02 mol.
Step 5: Calculate the concentrations of Ca2+ and SO2−
4ions in the resulting
solution. Total volume of the resulting solution = 100 mL + 150 mL = 250 mL
= 0.25 L
Concentration of Ca2+ =moles of Ca2+
total volume of solution =0.02 mol
0.25 L = 0.08 M
Concentration of SO2−
4=moles of SO2−
4
total volume of solution =0.02 mol
0.25 L = 0.08 M
Therefore, the concentrations of Ca2+ and SO2−
4ions in the resulting solution
are 0.08 M each.
Question 22
Question
A university laboratory needs to prepare a 250 mL solution that contains 0.10
M of sodium chloride (NaCl) and 0.20 M of silver nitrate (AgN O3). If these
two solutions are mixed together, how many grams of silver chloride (AgCl) will
precipitate out of the solution? Assume that the reaction goes to completion.
16
Solution
Step 1: Write the balanced chemical equation for the reaction between sodium
chloride and silver nitrate to form silver chloride:
NaCl(aq)+AgN O3(aq)→AgCl(s)+N aN O3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant initially. Given: Volume of NaCl solution = 250 mL = 0.250 L Molarity
of NaCl (M1) = 0.10 M Moles of N aCl:n1=M1×V= 0.10 mol/L×0.250 L=
0.025 mol
Molarity of AgNO3(M2) = 0.20 M Volume of AgN O3solution = 250 mL
= 0.250 L Moles of AgNO3:n2=M2×V= 0.20 mol/L ×0.250 L= 0.050 mol
Step 3: Determine the limiting reactant and calculate the moles of AgCl
formed. The balanced equation shows a 1:1 mole ratio between NaCl and AgCl.
Since the molar ratio between NaCl and AgCl is 1:1, the limiting reactant is
NaCl. Moles of AgCl formed: nAgCl =nN aCl = 0.025 mol
Step 4: Calculate the mass of AgCl formed using the molar mass of AgCl.
The molar mass of AgCl is calculated as the sum of the atomic masses of
silver (Ag) and chlorine (Cl): Ag = 107.87 g/mol,Cl = 35.45 g/mol AgCl =
107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed: mAgCl =nAgCl×MAgCl = 0.025 mol×143.32 g/mol =
3.58 g
Therefore, 3.58 grams of silver chloride will precipitate out of the solution.
Question 23
Question
A solution is prepared by mixing 120 mL of 0.3 M silver nitrate (AgNO3) with
80 mL of 0.2 M sodium chloride (NaCl). Calculate the mass of silver chloride
(AgCl) precipitate that forms in grams. The reaction is given by the equation:
AgNO3+ NaCl →AgCl + NaNO3
Solution
Step 1: Determine the limiting reagent by comparing the number of moles of
each reactant. Let’s start by calculating the number of moles for each reactant.
The number of moles can be calculated using the formula:
Number of moles = Molarity ×Volume (in L)
For silver nitrate (AgNO3): Number of moles = 0.3 mol/L ×0.12 L = 0.036
mol
For sodium chloride (NaCl): Number of moles = 0.2 mol/L ×0.08 L = 0.016
mol
17
Step 2: Determine the limiting reagent. Since the reaction stoichiometry is
1:1 between silver nitrate and sodium chloride, the limiting reagent is the one
that produces the fewer moles of AgCl. In this case, sodium chloride is the
limiting reagent because it produces only 0.016 mol of AgCl compared to the
0.036 mol produced by silver nitrate.
Step 3: Calculate the mass of silver chloride formed. The molar mass of
AgCl is:
1×Ag + 1 ×Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Using the number of moles of AgCl formed (0.016 mol) and its molar mass,
we can calculate the mass of AgCl formed:
0.016 mol ×143.32 g/mol = 2.293 g
Therefore, the mass of silver chloride precipitate formed is 2.293 grams.
Question 24
Question
Calculate the mass of lead(II) iodide (PbI2) that can be produced when 50.0
mL of a 0.200 M lead(II) nitrate solution is mixed with 75.0 mL of a 0.150 M
sodium iodide solution. Assume that the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and sodium iodide to form lead(II) iodide and sodium nitrate.
Pb(NO3)2(aq) + 2NaI(aq)→PbI2(s) + 2NaNO3(aq)
Step 2: Calculate the amount of lead(II) nitrate and sodium iodide used.
Amount of Pb(NO3)2= Volume ×Molarity
= 0.0500 L ×0.200 mol/L
= 0.0100 mol
Amount of NaI = Volume ×Molarity
= 0.0750 L ×0.150 mol/L
= 0.0113 mol
Step 3: Determine the limiting reactant and calculate the maximum amount
of lead(II) iodide that can be produced. Since lead(II) nitrate and sodium iodide
react in a 1:2 molar ratio, the limiting reactant is the lead(II) nitrate.
Moles of PbI2= 0.0100 mol Pb(NO3)2×1 mol PbI2
1 mol Pb(NO3)2
= 0.0100 mol
18
Step 4: Convert moles of lead(II) iodide to mass.
Mass of PbI2= Moles ×Molar Mass
= 0.0100 mol ×(207.2 g/mol)
= 2.07 g
Therefore, the mass of lead(II) iodide that can be produced is 2.07 grams.
Question 25
Question
Calculate the mass of lead(II) iodide (PbI2) that can be formed when 50.0 mL
of 0.100 M lead(II) nitrate (Pb(NO3)2) solution is mixed with 50.0 mL of 0.100
M potassium iodide (KI) solution. Assume the reaction goes to completion and
that lead(II) iodide is the only product.
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide.
Pb(NO3)2+ 2KI →PbI2+ 2KNO3
Step 2: Determine the limiting reactant by calculating the moles of each
reactant. For lead(II) nitrate:
moles = Volume ×Molarity
= (0.0500 L) ×(0.100 mol/L)
= 0.00500 mol
For potassium iodide:
moles = Volume ×Molarity
= (0.0500 L) ×(0.100 mol/L)
= 0.00500 mol
Both reactants have the same number of moles, so lead(II) nitrate is the
limiting reactant.
Step 3: Calculate the theoretical yield of lead(II) iodide by using the stoi-
chiometry from the balanced equation. From the balanced equation, 1 mole of
lead(II) nitrate produces 1 mole of lead(II) iodide.
moles of PbI2= moles of Pb(NO3)2= 0.00500 mol
19
Step 4: Convert moles of lead(II) iodide to mass using its molar mass.
Molar mass of PbI2= mass of Pb + 2 ×(mass of I)
= 207.2 g/mol + 2 ×126.9 g/mol
= 460.2 g/mol
Mass of PbI2= moles of PbI2×molar mass of PbI2
= 0.00500 mol ×460.2 g/mol
= 2.30 g
Therefore, the mass of lead(II) iodide that can be formed is 2.30 g.
Question 26
Question
Calculate the concentration of chloride ions in a solution that results from mix-
ing 100.0 mL of 0.200 M NaCl with 200.0 mL of 0.500 M MgCl2. Assume that
the volumes are additive.
Solution
Step 1: Find the moles of Cl−ions from NaCl. Given: Volume of NaCl solution,
V1= 100.0 mL = 0.100 L Concentration of NaCl, C1= 0.200 M
The moles of Cl−ions from NaCl can be calculated using the formula:
moles of ions = volume ×concentration
Therefore,
moles of Cl−from NaCl = 0.100 L ×0.200 M = 0.020 moles
Step 2: Find the moles of Cl−ions from MgCl2. Given: Volume of MgCl2
solution, V2= 200.0 mL = 0.200 L Concentration of MgCl2,C2= 0.500 M
The moles of Cl−ions from MgCl2can be calculated using the same formula:
moles of ions = volume ×concentration
Therefore,
moles of Cl−from MgCl2= 0.200 L ×0.500 M = 0.100 moles
Step 3: Find the total moles of Cl−ions. Since the volumes of the solutions
are additive, the total moles of Cl−ions is:
Total moles of Cl−= moles of Cl−from NaCl+moles of Cl−from MgCl2= 0.020 moles+0.100 moles = 0.120 moles
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Step 4: Calculate the concentration of chloride ions in the final solution. The
total volume of the final solution is the sum of the volumes of the two solutions:
Vtotal = 0.100 L + 0.200 L = 0.300 L
Therefore, the concentration of chloride ions in the final solution is:
Total moles of Cl−
Vtotal
=0.120 moles
0.300 L = 0.40 M
Question 27
Question
A solution is prepared by mixing 50.0 mL of 0.200 M silver nitrate (AgNO3)
with 50.0 mL of 0.100 M sodium chloride (NaCl). Determine the mass of silver
chloride (AgCl) that will precipitate.
Solution
Step 1: Write the balanced chemical equation for the reaction of silver nitrate
(AgNO3) with sodium chloride (NaCl) to form silver chloride (AgCl) precipitate:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant to find the maximum amount of
precipitate that can form. Calculating the moles of each reactant:
Moles of AgNO3= Volume (L) ×Molarity = 0.050 L ×0.200 mol/L =
0.010 mol
Moles of NaCl = Volume (L) ×Molarity = 0.050 L×0.100 mol/L = 0.005 mol
From the balanced chemical equation, the mole ratio of AgNO3to NaCl is
1:1. Since the moles of NaCl are less, it is the limiting reactant.
Step 3: Calculate the mass of silver chloride (AgCl) precipitate formed using
the moles of limiting reactant:
Moles of NaCl = 0.005 mol
Molar mass of AgCl: 1 atom of Ag + 1 atom of Cl = 107.87 g/mol +
35.45 g/mol = 143.32 g/mol
Mass of AgCl = Moles ×Molar mass = 0.005 mol ×143.32 g/mol = 0.717 g
Therefore, the mass of silver chloride (AgCl) that will precipitate is 0.717
grams.
Question 28
Question
Calculate the concentration of chloride ions in a solution prepared by mixing
100.0 mL of a 0.200 M solution of calcium chloride with 200.0 mL of a 0.500 M
solution of sodium chloride.
21
Solution
Step 1: Write the balanced chemical equation for the dissociation of calcium
chloride and sodium chloride in water.
CaCl2→Ca2+ + 2Cl−
NaCl →Na++ Cl−
Step 2: Calculate the moles of chloride ions from each solution. For calcium
chloride:
Moles of Cl−= Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.020 mol
For sodium chloride:
Moles of Cl−= Volume ×Molarity = 0.200 L ×0.500 mol/L = 0.100 mol
Step 3: Calculate the total moles of chloride ions in the final solution.
Total moles of Cl−= 0.020 mol + 0.100 mol = 0.120 mol
Step 4: Calculate the final volume of the solution.
Final volume = 100.0 mL + 200.0 mL = 300.0 mL = 0.300 L
Step 5: Calculate the concentration of chloride ions in the final solution.
Concentration of Cl−=Total moles of Cl−
Final volume =0.120 mol
0.300 L = 0.400 M
Therefore, the concentration of chloride ions in the final solution is 0.400 M.
Question 29
Question
A chemist needs to prepare 250 mL of a 0.05 M solution of silver nitrate
(AgNO3) by dissolving solid AgN O3in distilled water. The only source avail-
able is AgNO3solid and the chemist estimates the molar mass of AgNO3to
be 170.9 g/mol. What mass of AgNO3should the chemist dissolve in water to
prepare the desired solution?
Solution
Step 1: Calculate the number of moles of AgN O3required to make 250 mL of
a 0.05 M solution. Step 2: Use the molar mass of AgNO3to find the mass of
AgNO3required to prepare the solution.
Step 1: Given: Volume of solution to be prepared, V= 250 mL = 0.250 L
Molarity of the solution, M= 0.05 mol/L
22
Using the definition of molarity, which is given by M=n
V, where nis the
number of moles and Vis the volume in liters, we can rearrange the equation
to solve for n:
n=M×V n = 0.05 mol/L ×0.250 L n= 0.0125 mol
Therefore, the chemist needs 0.0125 moles of AgN O3to prepare the solution.
Step 2: Given: Molar mass of AgNO3,MM = 170.9 g/mol
To find the mass of AgNO3required, we can use the formula:
Mass = Number of moles ×Molar mass
Substitute the known values:
Mass = 0.0125 mol ×170.9 g/mol Mass = 2.13625 g
Therefore, the chemist should dissolve 2.13625 g of AgN O3in water to
prepare the desired 0.05 M solution.
Question 30
Question
A student is conducting an experiment to determine the concentration of chlo-
ride ions in a water sample. The student adds excess silver nitrate solution to
100.0 mL of the water sample, which precipitates all the chloride ions as sil-
ver chloride. After filtration, the mass of the silver chloride obtained is 0.563
g. Calculate the concentration of chloride ions in the original water sample in
units of mol/L.
(Silver chloride has a molar mass of 143.32 g/mol)
Solution
Step 1: Calculate the moles of silver chloride formed.
Moles of silver chloride = Mass of silver chloride
Molar mass =0.563 g
143.32 g/mol
Step 2: Calculate the moles of chloride ions in the silver chloride. 1 mole of
silver chloride contains 1 mole of chloride ions.
Moles of chloride ions = Moles of silver chloride
Step 3: Calculate the volume of the water sample in liters. Volume of water
sample = 100.0 mL = 0.100 L
Step 4: Calculate the concentration of chloride ions in the original water
sample.
Concentration of chloride ions = Moles of chloride ions
Volume of water sample
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Question 31
Question
Calculate the concentration of chloride ions in a solution when 100.0 mL of 0.200
M silver nitrate (AgNO3) is mixed with 100.0 mL of 0.150 M sodium chloride
(NaCl). The Ksp of silver chloride (AgCl) is 1.8×10−10.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant. Since both reactants have a 1:1
molar ratio in the balanced chemical equation, the reactant that produces less
silver chloride will be the limiting reactant.
Calculate the moles of AgNO3:
Moles of AgNO3= Volume ×Molarity
Moles of AgNO3= 0.100 L ×0.200 mol/L = 0.020 mol
Calculate the moles of NaCl:
Moles of NaCl = Volume ×Molarity
Moles of NaCl = 0.100 L ×0.150 mol/L = 0.015 mol
Since 0.015 moles of NaCl produces less silver chloride than 0.020 moles of
AgNO3, NaCl is the limiting reactant.
Step 3: Determine the moles of AgCl formed. From the balanced equation,
1 mole of NaCl forms 1 mole of AgCl. So, 0.015 moles of NaCl will form 0.015
moles of AgCl.
Step 4: Calculate the concentration of Cl−ions in the solution. The total
volume of the solution is 200.0 mL or 0.200 L. Therefore, the concentration of
Cl−ions in the solution is:
Concentration of Cl−=Moles of AgCl
Total Volume
Concentration of Cl−=0.015 mol
0.200 L = 0.075 mol/L
The concentration of chloride ions in the solution is 0.075 mol/L.
Question 32
Question
A solution contains 0.2 M of calcium chloride. If we add 0.1 M of sodium sulfate
to the solution, what is the concentration of calcium ions after precipitation
occurs? The Ksp of calcium sulfate is 2.4×10−5.
24
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between calcium chloride and sodium sulfate:
CaCl2(aq) + Na2SO4(aq)→CaSO4(s) + 2NaCl(aq)
Step 2: Calculate the initial concentration of calcium ions, Ca2+, beforetheprecipitationreactionoccurs :
[Ca2+]initial = 0.2 M
Step 3: Determine the concentration of calcium ions, Ca2+, aftertheprecipitationreactionoccurs.LetxbetheconcentrationofCa2+ionsatequilibrium.T hebalancedchemicalequationshowsa1 :
1molarratiobetweenCaCl2andCaSO4.T heref ore, theconcentrationof Ca2+ionsafterprecipitationis :
[Ca2+]final = 0.2 M + x
Step 4: Write the expression for the solubility product,
Ksp = [Ca2+][SO2−
4]
, and substitute the known values:
2.4×10−5= (0.2 + x)(0.1)
Step 5: Solve for x to find the concentration of calcium ions after precipita-
tion occurs:
2.4×10−5= 0.02 + 0.1x
0.1x= 2.4×10−5−0.02
x≈0.00016 M
Therefore, the concentration of calcium ions after precipitation occurs is
approximately 0.00016 M.
Question 33
Question
During a rainstorm, 2 inches of rain falls over a 1-acre area. If 1 inch of rain
is equivalent to 4.168 cubic feet of water, how many gallons of water fell on
the 1-acre area during the rainstorm? (Hint: 1 gallon is equivalent to 0.133681
cubic feet)
Solution
Step 1: Find the total volume of water that fell on the 1-acre area in cubic feet.
Since 1 inch of rain is equivalent to 4.168 cubic feet of water, the total volume
of water that fell on the 1-acre area can be calculated as follows:
2 inches ×4.168 cubic feet per inch = 8.336 cubic feet
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Step 2: Convert the total volume of water to gallons. Since 1 gallon is
equivalent to 0.133681 cubic feet, we can convert the total volume of water in
cubic feet to gallons as follows:
8.336 cubic feet
0.133681 cubic feet per gallon ≈62.37 gallons
Therefore, approximately 62.37 gallons of water fell on the 1-acre area during
the rainstorm.
Question 34
Question
A chemistry student is performing a precipitation reaction between aqueous
solutions of lead(II) nitrate and potassium iodide. If 50.0 mL of 0.200 M lead(II)
nitrate is mixed with excess potassium iodide, what mass of lead(II) iodide will
precipitate?
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and potassium iodide. The balanced equation is:
Pb(NO3)2(aq) + 2KI(aq) →PbI2(s) + 2KNO3(aq)
Step 2: Determine the limiting reactant by calculating the amount of lead(II)
iodide that would be formed from each reactant. Since lead(II) nitrate and
potassium iodide react in a 1:2 mole ratio, the moles of lead(II) nitrate is given
by:
moles = Molarity ×Volume = 0.200 mol/L ×0.0500 L = 0.0100 mol
Similarly, the moles of lead(II) iodide formed by lead(II) nitrate is:
moles PbI2= 0.0100 mol
Step 3: Calculate the mass of lead(II) iodide formed using the molar mass
of lead(II) iodide (461 g/mol):
Mass = mol ×Molar mass = 0.0100 mol ×461 g/mol = 4.61 g
Therefore, the mass of lead(II) iodide that will precipitate is 4.61 grams.
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Question 35
Question
A chemistry student is conducting an experiment where a silver nitrate solution
reacts with a sodium chloride solution to produce silver chloride precipitate.
The student mixes 100.0 mL of 0.200 M silver nitrate solution with 150.0 mL of
0.150 M sodium chloride solution. What mass of silver chloride precipitate will
form?
(Hint: Silver chloride is insoluble in water and will precipitate out of solu-
tion).
Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:
AgNO3+NaCl →AgCl +N aN O3
Step 2: Determine the limiting reactant by comparing the number of moles
of each reactant. The number of moles of silver nitrate:
moles of AgNO3= Molarity ×Volume (L)
moles of AgNO3= 0.200 mol/L ×0.100 L = 0.0200 mol
The number of moles of sodium chloride:
moles of NaCl = Molarity ×Volume (L)
moles of NaCl = 0.150 mol/L ×0.150 L = 0.0225 mol
Since silver nitrate and sodium chloride react in the ratio of 1:1, the limiting
reactant is silver nitrate because it produces fewer moles of product.
Step 3: Calculate the mass of silver chloride precipitate that will form. The
molar mass of silver chloride (AgCl) is approximately 143.32 g/mol.
moles of AgCl = moles of AgNO3
mass of AgCl = moles of AgCl ×molar mass of AgCl
mass of AgCl = 0.0200 mol ×143.32 g/mol = 2.8664 g
Therefore, the mass of silver chloride precipitate that will form is 2.8664
grams.
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