CHEM 301 - ORGANIC CHEMISTRY
I - Solubility and polarity calculations
Question Bank - Set 3
Liberty University
Question 1
Question
Predict the solubility of each of the following compounds in water: (i) 1,2-
dichloroethane (ii) ethyl acetate (iii) 1-octanol
Solution
In order to predict the solubility of each compound in water, we need to consider
the polarity of the compounds and the presence of any functional groups that
can participate in hydrogen bonding with water molecules.
Step 1: Evaluate the polarity of each compound. (i) 1,2-dichloroethane is
nonpolar due to the symmetrical distribution of the two chlorine atoms can-
celling out any dipole moment. (ii) ethyl acetate is polar due to the presence
of the ester functional group, which contains a polar carbonyl group and an
oxygen atom that can participate in hydrogen bonding. (iii) 1-octanol is polar
due to the hydroxyl group, which can participate in hydrogen bonding.
Step 2: Determine the solubility in water based on polarity. (i) 1,2-
dichloroethane is nonpolar and therefore insoluble in water. (ii) ethyl acetate is
polar and can form hydrogen bonds with water, so it is soluble in water. (iii)
1-octanol is polar and can form hydrogen bonds with water, so it is also soluble
in water.
Therefore, the predicted solubility of each compound in water is: (i) 1,2-
dichloroethane: Insoluble (ii) ethyl acetate: Soluble (iii) 1-octanol: Soluble
Question 2
Question
Determine which of the following pairs of compounds would be soluble in water,
and which would be more polar:
Compound A: Cyclohexanol Compound B: Octane
Solution
Step 1: Solubility in Water - Cyclohexanol has a hydroxyl group (−OH) which
can participate in hydrogen bonding with water molecules, making it soluble in
water. - Octane is a nonpolar hydrocarbon which does not have any functional
groups that can form significant intermolecular interactions with water, making
it insoluble in water.
Step 2: Polarity - Cyclohexanol is polar due to the presence of the hydroxyl
group, which can participate in hydrogen bonding. - Octane is nonpolar because
it consists of only carbon and hydrogen atoms and lacks any polar functional
groups.
Therefore, Compound A (Cyclohexanol) would be soluble in water and more
polar than Compound B (Octane).
Question 3
Question
Calculate the solubility (in g/L) of benzoic acid in water at 25
°
C. The solubility
of benzoic acid in water is 3.5 g/100 mL at that temperature.
Solution
Step 1: Start by converting the given solubility from g/100 mL to g/L.
Solubility in g/L = 3.5 g
100 mL×1000 mL
1 L
= 35 g/L
Step 2: Use the solubility of benzoic acid in water to perform the final
calculation.
Solubility in water = 35 g/L
Therefore, the solubility of benzoic acid in water at 25
°
C is 35 g/L.
2
Question 4
Question
Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. Given that
the solubility product constant, Ksp, for benzoic acid is 1.7×10−3at this
temperature.
Solution
Step 1: Write the dissociation equation for benzoic acid dissolving in water:
C7H6O2(s) ⇌C7H6O2(aq)
The solubility equilibrium expression can be given as:
Ksp = [C7H6O2]2
Step 2: Let x be the molar solubility of benzoic acid. Therefore, the equi-
librium concentrations are:
[C7H6O2] = x
Substitute the equilibrium concentrations into the solubility product con-
stant expression:
1.7×10−3= (x)2
x=p1.7×10−3
x= 0.0412 M
Therefore, the solubility of benzoic acid in water at 25◦C is 0.0412 M.
Question 5
Question
A student is given two organic compounds, compound A and compound B, and
is asked to determine which one is more soluble in water. Compound A has
a molecular weight of 150 g/mol and a log P value of 2.3, while compound B
has a molecular weight of 250 g/mol and a log P value of 1.8. Based on these
properties, which compound is expected to be more soluble in water? Justify
your answer.
Solution
To determine which compound is more soluble in water, we can use the par-
tition coefficient (P = solubility in octanol
solubility in water ) as a measure of how hydrophilic or
hydrophobic a compound is.
3
Step 1: Calculate the partition coefficient for compound A. The
partition coefficient (P) can be calculated using the formula:
P = 10log P
For compound A:
PA= 102.3≈199.53
Step 2: Calculate the partition coefficient for compound B. For
compound B:
PB= 101.8≈63.10
Step 3: Determine the solubility in water. Since a higher partition
coefficient indicates greater hydrophobicity, compound B is expected to be less
soluble in water compared to compound A. Therefore, compound A is expected
to be more soluble in water despite its lower molecular weight.
Question 6
Question
Calculate the solubility of compound X in water at 25
°
C. The partition coeffi-
cient of X between water and an organic solvent is 0.15. Assume the volume
change upon dissolution is negligible.
Solution
Step 1: Write the partition coefficient equation Step 2: Write the equation for
solubility of X in water Step 3: Calculate the solubility of X in water
Step 1: The partition coefficient (KOW) is defined as the ratio of solute
concentration in an organic solvent to the solute concentration in water. Math-
ematically,
KOW =[X in organic solvent]
[X in water] = 0.15
Step 2: Let’s assume the initial amount of compound X is x. If the solubility
of X in water is smol/L, then the solubility in the organic solvent will be x−s
mol/L. Using the partition coefficient equation, we have
0.15 = x−s
s
Step 3: Solving for s,
0.15s=x−s
0.15s+s=x
s(0.15 + 1) = x
s=x
1.15
4
Since the volume change upon dissolution is negligible, the moles of X before
and after dissolution are the same. Therefore, x= solubility of X in water at
25
°
C.
Thus, the solubility of compound X in water at 25
°
C is x
1.15 =solubility of X in organic solvent
1.15 .
Question 7
Question
Calculate the solubility of benzoic acid in water at 25◦C. The solubility of ben-
zoic acid in water is 3.4 g/L at 100◦C and its partition coefficient between water
and benzene at room temperature is 10.5. Assume the density of water is 1.0
g/mL and benzene is 0.88 g/mL at room temperature.
Solution
Step 1: Calculate the partition coefficient Kat 25◦C using the given value at
another temperature. Given that T1= 100◦C, T2= 25◦C, K1= 10.5, and
∆Hsol =−33 kJ/mol, we can use the van’t Hoff equation:
ln K2
K1=−∆Hsol
R1
T2
−1
T1
where R= 8.314 J/mol ·K=0.008314 kJ/mol ·K.
ln K
10.5=−
−33
0.008314 1
298 −1
373
ln K
10.5= 14.2581
K
10.5=e14.2581
K= 10.5×e14.2581
K≈2.04 ×106
Step 2: Calculate the molar solubility of benzoic acid in water. Given that
K= 2.04 ×106, let the molar solubility of benzoic acid in water at 25◦C be x
mol/L. The equilibrium between the solid benzoic acid and the dissolved benzoic
acid is represented by:
C6H5COOH(s)⇌C6H5COOH(aq)
The equilibrium constant Keq is given by:
Keq =[C6H5COOH(aq)]
[C6H5COOH(s)] ≈x
3.4= 2.04 ×106
5
Solving for x:
x= 2.04 ×106×3.4
x≈6.94 ×106mol/L
Step 3: Calculate the mass solubility of benzoic acid in water. Given that the
molar mass of benzoic acid is 122.12 g/mol, we can convert the molar solubility
to mass solubility:
Mass solubility = 6.94 ×106mol/L ×122.12 g/mol
Mass solubility ≈8.47 ×105g/L
Thus, the solubility of benzoic acid in water at 25◦C is approximately 847 g/L.
Question 8
Question
A student is trying to determine the solubility of compound X in different sol-
vents. Compound X is known to be nonpolar and has a molecular weight of 150
g/mol. The student finds that compound X is soluble in hexane but insoluble
in water. Based on this information, calculate the solubility (in grams per liter)
of compound X in hexane.
(Hint: Use the molecular weight of the compound to calculate the molarity
in hexane, and then convert this to solubility in grams per liter.)
Solution
Step 1: Calculate the molarity of compound X in hexane
The molarity of compound X in hexane can be calculated as:
Molarity = grams of solute
molecular weight of solute ×volume of solvent (L)
Since the molecular weight of compound X is 150 g/mol, we can use this to
calculate the molarity in hexane.
Step 2: Convert the molarity to solubility in grams per liter
Given that the student found compound X to be soluble in hexane, the
calculated molarity can be converted to solubility in grams per liter using the
relationship:
Solubility (g/L) = Molarity ×Molecular weight of solute
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Question 9
Question
A student is conducting a solubility experiment in which they dissolve 0.050
moles of a nonpolar organic compound in 50 mL of water at 25
°
C. The student
notices that the compound partially dissolves. Calculate the solubility of the
compound in water in mol/L. Assume the volume of the organic compound is
negligible.
Solution
Step 1: Calculate the molarity of the water solution. Given: - Moles of organic
compound = 0.050 mol - Volume of water = 50 mL = 0.050 L
The molarity of the water solution is calculated using the formula:
Molarity =Number of moles
Volume in liters
Plugging in the values:
Molarity =0.050 mol
0.050 L
Molarity = 1.0 mol/L
Step 2: Calculate the solubility of the compound in water. Since the com-
pound partially dissolves, let’s assume x moles of the compound dissolves in 1
L water. The equilibrium expression for the dissolution of the compound is:
Compound ⇌(aq)
The solubility product expression is:
Ksp = [Compound] = x
The solubility product constant (Ksp) for a nonpolar compound is assumed
to be very small. Since the compound partially dissolves, we can assume that
the equilibrium concentration of the compound is much less than 1.0 mol/L.
Therefore, the solubility of the compound in water at 25
°
C is very low, close
to or less than 1.0 mol/L.
Question 10
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The Ksp
of benzoic acid in water is 8.3×10−5. Assume the density of water is 1.0 g/mL.
7
Solution
Step 1: Write the dissociation reaction of benzoic acid in water:
C6H5COOH ⇌C6H5COO−+ H+
Step 2: Construct the equilibrium expression for the dissociation of benzoic
acid:
Ksp = [C6H5COO−][H+]
Step 3: Let x be the molar solubility of benzoic acid. Since benzoic acid is a
weak acid, we can approximate the equilibrium concentrations of C6H5COO−
and H+as x. Therefore, the expression becomes:
Ksp = (x)(x)
Ksp =x2
Step 4: Substitute the given Ksp value into the equilibrium expression and
solve for x:
8.3×10−5=x2
x=p8.3×10−5
x≈0.0091 M
Step 5: Calculate the solubility of benzoic acid in g/L:
Molar mass of benzoic acid = (6×12.01)+(5×1.01)+12.01+16.00+1.01 = 122.12 g/mol
Solubility in g/L = (0.0091 mol/L) ×(122.12 g/mol) ×(1 L/1000 mL)
Solubility in g/L ≈0.0111 g/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0111 g/L.
Question 11
Question
A student needs to determine the solubility of a compound in water at room
temperature. The student knows that the compound is a carboxylic acid and
has a molecular weight of 122 g/mol. The student performed a solubility test
and found that 0.5 g of the compound dissolves in 10 mL of water. Calculate
the solubility of the compound in mol/L.
8
Solution
Step 1: Calculate the molar mass of the compound. Step 2: Determine the
number of moles of the compound that dissolve in 10 mL of water. Step 3:
Convert the volume of water to liters. Step 4: Calculate the solubility of the
compound in mol/L.
Step 1: The molar mass of the compound is 122 g/mol.
Step 2: The number of moles of the compound that dissolve in 10 mL of
water can be calculated using the given mass and molar mass. Number of moles
= Mass / Molar mass Number of moles = 0.5 g / 122 g/mol
Step 3: Convert the volume of water from milliliters to liters. 1 mL = 0.001
L Volume of water = 10 mL * 0.001 L/mL
Step 4: Now, calculate the solubility of the compound in mol/L. Solubility
(in mol/L) = Number of moles / Volume of water
Therefore, the solubility of the compound in water at room temperature is
0.5 g/122 g/mol
10 mL ×0.001 L/mL
mol/L.
Question 12
Question
A student is given a sample of compound X and is asked to determine its solu-
bility in water. After running solubility tests, the student finds that compound
X is soluble in both water and hexane. Based on this information, determine
the possible functional groups present in compound X and explain the reasoning
behind these possibilities.
Solution
Step 1: Since compound X is soluble in water, it must possess some polar func-
tional groups that can interact with water molecules through hydrogen bonding
or dipole-dipole interactions.
Step 2: The solubility of compound X in hexane suggests the presence of
nonpolar functional groups that can interact with nonpolar solvents like hexane
through London dispersion forces.
Step 3: Keeping in mind the solubility information in both water and hexane,
the possible functional groups in compound X could include alcohols, amines,
carboxylic acids, and esters.
Step 4: Alcohols have a polar hydroxyl (-OH) functional group that allows
them to form hydrogen bonds with water molecules, making them soluble in
water. However, the nonpolar alkyl group in alcohols also allows them to dissolve
in nonpolar solvents like hexane.
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Step 5: Amines contain a polar amino (-NH2) functional group, which can
form hydrogen bonds with water molecules, leading to their solubility in water.
The alkyl group in amines contributes to their solubility in nonpolar solvents
like hexane.
Step 6: Carboxylic acids have a polar carboxyl (-COOH) group that allows
them to form hydrogen bonds with water, making them soluble in water. The
nonpolar alkyl group in carboxylic acids enables them to dissolve in nonpolar
solvents like hexane.
Step 7: Esters contain a polar carbonyl (C=O) group and can form hydrogen
bonds with water molecules, leading to their solubility in water. The nonpolar
alkyl groups in esters contribute to their solubility in nonpolar solvents like
hexane.
Step 8: Therefore, compound X could potentially contain alcohols, amines,
carboxylic acids, or esters as functional groups based on its solubility charac-
teristics in both water and hexane.
Question 13
Question
A student is given a white solid compound to identify in the laboratory. The
compound is sparingly soluble in water but dissolves readily in ether. Upon
adding HCl to the white solid, a white precipitate is formed. On the basis of
the solubility properties described, which functional group or groups are most
likely present in the compound?
Solution
Step 1: The compound is sparingly soluble in water but dissolves readily in
ether. This information suggests that the compound is nonpolar or has a low
polarity.
Step 2: The compound dissolves in ether, a nonpolar solvent, indicating that
the compound is likely nonpolar or has a low polarity. Nonpolar compounds
tend to dissolve well in nonpolar solvents such as ether.
Step 3: Upon addition of HCl, a white precipitate is formed. This suggests
the presence of a functional group that can react with HCl to form an insoluble
product.
Step 4: The functional group that reacts with HCl to form an insoluble
product is likely an amine group (RNH2, R2NH, or R3N). Amines react with
HCl to form ammonium chloride salts, which are typically insoluble in water
and precipitate out.
Step 5: Therefore, the functional group most likely present in the compound
based on the given solubility properties is an amine group.
10
Question 14
Question
An organic compound, X, has a molecular weight of 86 g/mol. When 0.5 g of
compound X is dissolved in 10 mL of water at 25
°
C, it forms a clear solution.
However, when 0.5 g of compound X is dissolved in 10 mL of diethyl ether at
25
°
C, it forms a cloudy solution. Determine whether compound X is more likely
to be polar or nonpolar.
Solution
Step 1: Calculate the molarity of compound X in water solution. Given: - Mass
of X = 0.5 g - Volume of water = 10 mL = 0.01 L - Molecular weight of X =
86 g/mol
First, calculate the number of moles of X:
Moles of X = Mass of X
Molecular weight of X =0.5 g
86 g/mol
Moles of X ≈0.00581 mol
Next, calculate the molarity of X in water:
Molarity = Moles of X
Volume of solution =0.00581 mol
0.01 L
Molarity = 0.581 M
Step 2: Calculate the molarity of compound X in diethyl ether solution.
Given: - Volume of diethyl ether = 10 mL = 0.01 L
Since the compound X forms a cloudy solution in diethyl ether, it is less
soluble in diethyl ether than in water. This indicates that compound X is more
likely to be polar due to the polarity of water.
Thus, compound X is more likely to be a polar compound.
Question 15
Question
Calculate the solubility of compound X in water at 25
°
C. The observed solubility
is 0.04 mol/L. The partition coefficient of X between water and benzene is 10.
Solution
Step 1: Calculate the solubility of compound X in benzene. Given: Partition
coefficient (Kow) = 10 Let the solubility of X in benzene be xmol/L.
11
The partition coefficient is defined as:
Kow =Concentration of X in benzene
Concentration of X in water
10 = x
0.04
x= 0.4 mol/L
Therefore, the solubility of compound X in benzene is 0.4 mol/L.
Step 2: Calculate the solubility of X in water. Let the solubility of X in
water be ymol/L.
The observation is that the solubility of X in water is 0.04 mol/L, and the
solubility in benzene is 0.4 mol/L. This means the solute prefers to dissolve in
benzene over water.
Hence, y= 0.04 mol/L.
Therefore, the solubility of compound X in water at 25
°
C is 0.04 mol/L.
Question 16
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The Ksp
for benzoic acid at 25
°
C is 1.3×10−3.
Solution
Step 1: Write the dissociation equilibrium equation for benzoic acid:
C6H5COOH ⇌C6H5COO−+ H+
Step 2: Write the expression for the Ksp of benzoic acid:
Ksp = [C6H5COO−][H+]
Step 3: Let x be the solubility of benzoic acid. Since benzoic acid is a weak
acid, its solubility is equal to its dissociation in water as per the equation in
Step 1:
C6H5COOH(s)⇌C6H5COO−(aq)+H+(aq)
The equilibrium concentrations will be: [C6H5COO−] = xand [H+] = x
Step 4: Substitute the equilibrium concentrations into the Ksp expression:
Ksp =x×x=x2
Step 5: Plug in the given Ksp value:
1.3×10−3=x2
12
Step 6: Solve for x:
x=p1.3×10−3
x≈0.0361 M
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0361 M.
Question 17
Question
A student is conducting an experiment in which they need to dissolve 0.10 moles
of compound X in 500 mL of water. Compound X has a molar mass of 120 g/mol
and a solubility in water of 0.5 g/L. Is it possible to dissolve all 0.10 moles of
compound X in 500 mL of water? If not, calculate the maximum amount of
compound X that can be dissolved in the water.
Solution
Step 1: Calculate the maximum amount of compound X in grams that can be
dissolved in 500 mL of water. Given: - Molar mass of compound X = 120 g/mol
- Solubility of compound X in water = 0.5 g/L - Number of moles of compound
X to be dissolved = 0.10 moles - Volume of water available = 500 mL
First, convert 500 mL to liters:
500 mL = 500 ×10−3L=0.500 L
To find the maximum amount of compound X that can be dissolved in water,
we can use the solubility information:
Maximum amount of compound X = Solubility ×Volume of water
Maximum amount of compound X = 0.5 g/L ×0.500 L
Maximum amount of compound X = 0.25 g
Step 2: Calculate the amount of compound X in grams that corresponds to
0.10 moles. Given: - Number of moles of compound X to be dissolved = 0.10
moles
To find the amount of compound X in grams corresponding to 0.10 moles,
we can use the molar mass:
Amount of compound X for 0.10 moles = Number of moles ×Molar mass
Amount of compound X for 0.10 moles = 0.10 moles ×120 g/mol
Amount of compound X for 0.10 moles = 12 g
Since the amount of compound X for 0.10 moles (12 g) is greater than the
maximum amount that can be dissolved (0.25 g), it is not possible to dissolve
all 0.10 moles of compound X in 500 mL of water. The maximum amount that
can be dissolved is 0.25 g.
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Question 18
Question
A student is given two compounds, compound A and compound B, and is asked
to predict which compound is more soluble in water based on their structures.
Compound A has a molecular weight of 150 g/mol and consists of a long carbon
chain with a hydroxyl group (-OH) at one end. Compound B has a molecular
weight of 120 g/mol and consists of a cyclic structure with a carbonyl group
(C=O). Calculate the HLB (hydrophilic-lipophilic balance) value for each com-
pound and determine which compound is more likely to be soluble in water.
Solution
Step 1: Calculate the HLB value for compound A.
HLBA = 20 ×nOH
MW + 16 ×nCOOH
MW + 1.8×nester
MW
Given that compound A has a molecular weight of 150 g/mol and contains
one hydroxyl group (-OH):
HLBA = 20 ×1
150=20
150 = 0.133
Step 2: Calculate the HLB value for compound B.
HLBB = 20 ×nOH
MW + 16 ×nCOOH
MW + 1.8×nester
MW
Given that compound B has a molecular weight of 120 g/mol and contains
one carbonyl group (C=O):
HLBB = 16 ×1
120=16
120 = 0.133
Step 3: Compare the HLB values for compound A and compound B.
Since both compound A and compound B have the same HLB value of
0.133, this means that they both have similar hydrophilic-lipophilic balance
values. Therefore, based solely on the HLB values, it is difficult to predict
which compound is more likely to be soluble in water. Other factors such as the
specific interactions between the compounds and water molecules would need
to be considered to determine solubility.
Question 19
Question
An organic compound X has a molecular weight of 150 g/mol and an experi-
mentally determined solubility of 5 g/L at 25
°
C. Determine the solubility of X
in hexane (C6H14) at the same temperature. The density of hexane is 0.659
g/mL.
14
Solution
Step 1: Calculate the molar solubility of compound X in water. Given that the
solubility of X in water is 5 g/L and its molecular weight is 150 g/mol, we can
calculate the molar solubility in water as follows:
Molar solubility in water = 5 g/L
150 g/mol =1
30 mol/L
Step 2: Calculate the partition coefficient, P, of compound X between water
and hexane. The partition coefficient is defined as the ratio of the solubility of
a compound in two immiscible solvents.
P=Solubility of X in hexane
Solubility of X in water
Step 3: Using the partition coefficient to find the solubility of X in hexane.
We know that the solubility of X in water is 5 g/L and the molar solubility in
water is 1
30 mol/L. Let the solubility of X in hexane be xg/L. From the partition
coefficient formula, we have:
P=x
5=1
30
Solving for x, we get:
x=5
30 =1
6g/L
Therefore, the solubility of compound X in hexane at 25
°
C is 1
6g/L.
Question 20
Question
A student wants to separate a mixture of benzoic acid (solubility in water at
20
°
C is 3.5 g/L) and biphenyl (solubility in water at 20
°
C is 0.02 g/L) using the
principle of solubility. The student knows that the two compounds have different
solubilities in water due to the difference in polarity. Discuss the solubility of
benzoic acid and biphenyl in water based on their polarity.
Solution
Step 1: Benzoic acid is a polar molecule due to the presence of a carboxylic
acid group, which allows it to form hydrogen bonds with water molecules. This
results in higher solubility in water compared to nonpolar biphenyl.
Step 2: Biphenyl is a nonpolar molecule because it consists of two benzene
rings connected by a single bond. The lack of polar functional groups makes
biphenyl insoluble or poorly soluble in water.
Step 3: In the separation process, the student can exploit the difference in
solubility to dissolve benzoic acid in water and leave biphenyl undissolved. This
15
can be achieved by selecting an appropriate solvent that will dissolve benzoic
acid but not biphenyl, allowing for the separation of the two compounds based
on their polarity.
Step 4: By understanding the solubility and polarity of organic compounds,
the student can effectively design a separation strategy for mixtures containing
both polar and nonpolar compounds.
Question 21
Question
A student is conducting an experiment in the laboratory and produces a solution
by mixing 50.0 mL of water and 50.0 mL of diethyl ether. The student wants
to know if the resulting solution is homogeneous or heterogeneous based on its
solubility and polarity. Given that water has a density of 1.00 g/mL and diethyl
ether has a density of 0.71 g/mL, calculate the solubility parameter () for water
and diethyl ether. Based on the solubility parameters, determine if the solution
is likely to be homogeneous or heterogeneous.
Solution
Step 1: Calculate the solubility parameter () for water and diethyl ether using
the equation:
δ=qδ2
d+δ2
p
where δdis the dispersion component and δpis the polar component of the
solubility parameter.
For water:
δd,water = 14.6 MPa0.5
δp,water = 20.0 MPa0.5
δwater =p14.62+ 20.02= 25.0 MPa0.5
For diethyl ether:
δd,ether = 14.5 MPa0.5
δp,ether = 5.8 MPa0.5
δether =p14.52+ 5.82= 15.6 MPa0.5
Step 2: Compare the solubility parameters of water and diethyl ether. Since
the solubility parameters of water (25.0 MPa0.5) and diethyl ether (15.6 MPa0.5)
are significantly different, the solution is likely to be heterogeneous. Water,
being more polar than diethyl ether, will not dissolve diethyl ether completely,
resulting in a heterogeneous solution.
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Question 22
Question
Indicate whether each of the following compounds is more soluble in water or
in hexane based on their polarity:
A) Butanol (CHOH)
B) Pentanoic acid (CHCOOH)
C) Toluene (CH)
D) Acetone (CHCOCH)
Solution
Step 1: Determine the polarity of each compound based on its functional groups.
A) Butanol: contains a hydroxyl (-OH) group, making it a polar com-
pound.
B) Pentanoic acid: contains a carboxyl (-COOH) group, making it a polar
compound.
C) Toluene: a nonpolar compound as it consists of only carbon and hy-
drogen atoms.
D) Acetone: a polar compound due to the presence of the carbonyl group.
Step 2: Evaluate the solubility of each compound in water or hexane based
on their polarity.
A) Butanol: more soluble in water due to its polar nature and ability to
form hydrogen bonds with water molecules.
B) Pentanoic acid: soluble in water due to the polar carboxyl group that
can interact with water molecules through hydrogen bonding.
C) Toluene: more soluble in hexane, a nonpolar solvent, as both are non-
polar compounds that can interact through London dispersion forces.
D) Acetone: soluble in both water and hexane, but more soluble in water
due to its polar nature and ability to form hydrogen bonds with water
molecules.
Therefore, the solubility of the compounds in either water or hexane can
be predicted based on their polarity and ability to interact with the solvent
molecules.
17
Question 23
Question
Determine which compound, A or B, would be more soluble in water based on
their respective chemical structures and polarities. Justify your answer.
Compound A: 1-hexanol (CH3(CH2)5OH) Compound B: 1-hexanone (CH3(CH2)4COCH3)
Solution
Step 1: Analyze the chemical structures of the compounds.
Compound A (1-hexanol) contains a hydroxyl group (OH), making it a polar
molecule. The compound exhibits a polar covalent bond between the carbon
and oxygen atoms of the hydroxyl group.
Compound B (1-hexanone) contains a carbonyl group (CO), resulting in
a polar molecule due to the electronegativity difference between carbon and
oxygen in the carbonyl group.
Step 2: Determine the solubility of each compound in water based on polar-
ity.
Water is a polar molecule due to its bent structure and unequal sharing
of electrons between oxygen and hydrogen atoms. Polar solutes are generally
soluble in polar solvents like water.
Since both compounds A and B are polar molecules, both are expected to be
soluble in water due to the ability of polar solutes to dissolve in polar solvents.
However, the presence of an additional hydrogen bonding site in compound A
(OH group) compared to compound B (carbonyl group) may lead to stronger
interactions with water molecules, potentially making compound A more soluble
in water than compound B. Thus, compound A (1-hexanol) is expected to be
more soluble in water than compound B (1-hexanone).
Question 24
Question
An organic compound with the molecular formula C6H12Ois found to be soluble
in water. However, it is insoluble in hexane (C6H14), a nonpolar solvent. Pro-
vide a possible structure for this compound and explain its solubility behavior
in terms of its polarity.
Solution
Step 1: Calculate the degree of unsaturation (DU) to determine the compound’s
possible structures. The formula for degree of unsaturation is:
DU =2C+ 2 −H+X−N
2
18
Where: C= 6 (number of carbon atoms), H= 12 (number of hydrogen
atoms), and X= 16 (number of halogen atoms).
Since there are no halogens or nitrogen in the formula, we have:
DU =2∗6+2−12
2= 1
This indicates that the compound is likely to contain a double bond or a
ring.
Step 2: Given that the compound is soluble in water but insoluble in hexane,
it suggests that the compound is polar. Water is a polar solvent, while hexane
is nonpolar. A possible structure that fits the molecular formula C6H12Oand
is polar is a cyclic ether:
CH3−CH2−CH2−O−CH2−CH2−CH3
The presence of the oxygen atom in the ether functional group makes the
molecule polar. The oxygen atom is more electronegative than carbon and
hydrogen, creating a partial negative charge on the oxygen and partial positive
charges on the carbons. This polarity allows the compound to form hydrogen
bonds with water molecules, making it soluble in water. However, the nonpolar
hexane cannot form the necessary interactions to dissolve the compound.
Question 25
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The
given Ksp value for benzoic acid is 1.7×10−3mol/L.
Solution
Step 1: Write the dissociation equation for benzoic acid in water:
C6H5COOH ⇌C6H5COO−+H+
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [C6H5COO−][H+]
Step 3: Since benzoic acid is a weak acid, we need to assume that there
is negligible dissociation of C6H5COOH (essentially none). Therefore, we can
assume that the concentrations of C6H5COO−and H+are equal to the initial
solubility of benzoic acid. Let’s denote the solubility of benzoic acid as x.
Step 4: Substitute the expressions for the concentrations into the Ksp ex-
pression:
1.7×10−3=x·x
1.7×10−3=x2
19
Step 5: Solve for xto find the solubility of benzoic acid:
x=p1.7×10−3
x≈0.0412 mol/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0412 mol/L.
Question 26
Question
A student is performing a solubility test for a compound and observes that it
dissolves in ether but not in water. The student also determines the compound’s
partition coefficient between water and ether to be 3.2. Calculate the percentage
of the compound that would dissolve in ether if 10 g of the compound were added
to 100 mL of ether.
Solution
Step 1: Calculate the mass of the compound that would dissolve in water and in
ether. Let the mass of the compound that dissolves in water be xg. Therefore,
the mass of the compound that dissolves in ether would be 10 −xg.
Step 2: Use the definition of the partition coefficient to set up an equation.
Kp=[Compound in ether]
[Compound in water],
where Kp= 3.2.
Step 3: Express the concentrations in terms of masses and volumes. Given
that the compound is added to 100 mL of ether: - The concentration of the
compound in ether is 10 −x
100 g/mL. - The concentration of the compound in
water is x
100 g/mL.
Step 4: Substitute the concentrations into the partition coefficient equation.
3.2 = 10 −x
x.
Step 5: Solve the equation for x.
3.2x= 10 −x
4.2x= 10
x=10
4.2≈2.38 g
20
Thus, approximately 2.38 g of the compound dissolves in water, and 10 −
2.38 = 7.62 g of the compound dissolves in ether.
Step 6: Calculate the percentage of the compound that dissolves in ether.
Percentage in ether = Mass in ether
Total mass ×100
Percentage in ether = 7.62
10 ×100
Percentage in ether ≈76.2%
Therefore, approximately 76.2
Question 27
Question
Calculate the solubility of benzoic acid (C7H6O2) in water at 25
°
C. The solu-
bility product constant for benzoic acid in water is 6.2×10−3M2.
Solution
Step 1: Write the equilibrium equation representing the dissolution of benzoic
acid in water. Step 2: Write the expression for the solubility product constant
(Ksp). Step 3: Define the variables and set up the equilibrium expression for
benzoic acid. Step 4: Solve for the molar solubility of benzoic acid in water.
Question 28
Question
A compound has a solubility of 0.15 g/100 mL in water at 30
°
C. Calculate the
solubility product constant (Ksp) for this compound in water at 30
°
C. Assume
the compound completely dissociates in water.
Solution
Step 1: Calculate the molar solubility of the compound.
Molar solubility = Solubility (g/L)
Molar mass (g/mol)
Molar solubility = 0.15 g/100 mL ×10 mL/L
M
Molar solubility = 0.015 g/L
M
21
Step 2: Calculate the Ksp using the molar solubility. The compound com-
pletely dissociates into its ions. If the compound is represented as AxBy, the
dissociation equation would be AxBy⇌xAz++yBw−. Then, the Ksp expres-
sion is:
Ksp = [Az+]x[Bw−]y
Since there is complete dissociation, we can say that:
[Az+] = x×Molar solubility = x×0.015 g/L
M
[Bw−] = y×Molar solubility = y×0.015 g/L
M
Substitute these expressions into the Ksp expression:
Ksp =x×0.015 g/L
Mxy×0.015 g/L
My
Question 29
Question
Compound A has a solubility of 0.15 g/L in water at 25
°
C. Compound B has a
solubility of 0.30 g/L in water at 25
°
C. Which compound is more polar? Justify
your answer with calculations.
Solution
Step 1: Calculate the molar solubility of each compound. Let’s assume the
molar mass of compound A is MAg/mol and the molar mass of compound B
is MBg/mol.
The molar solubility of compound A (SA) can be calculated as:
SA=0.15 g/L
MAg/mol
The molar solubility of compound B (SB) can be calculated as:
SB=0.30 g/L
MBg/mol
Step 2: Compare the molar solubilities of compounds A and B. To compare
the polarities of compounds A and B, we can compare their molar solubilities.
Higher molar solubility indicates a more polar compound.
If SA> SB, then compound A is more polar. If SA< SB, then compound
B is more polar.
Step 3: Use experimental data to compare the polarities of compounds A
and B. Assume MA= 100 g/mol and MB= 150 g/mol for the purpose of
comparison.
22
Calculate the molar solubility of compound A:
SA=0.15 g/L
100 g/mol = 0.0015 mol/L
Calculate the molar solubility of compound B:
SB=0.30 g/L
150 g/mol = 0.0020 mol/L
Step 4: Compare the molar solubilities of compounds A and B: Since SB>
SA, compound B is more polar than compound A based on their experimental
solubilities.
Question 30
Question
A student is trying to decide which solvent to use in a recrystallization procedure
for a compound. Compound A has a melting point of 85
°
C and is known to
be soluble in both ethanol and water. The student knows that the solubility of
a compound is related to the polarity of the solvent, with more polar solvents
dissolving more polar compounds. The student also knows that the melting
point of a compound is related to its purity, with impure compounds having
lower melting points. Which solvent, water or ethanol, would be more suitable
for recrystallizing compound A to obtain the purest crystals? Justify your
answer.
Solution
Step 1: Determine the polarity of compound A in relation to water and ethanol.
- Water is a highly polar solvent due to its ability to hydrogen bond. - Ethanol
is also a polar solvent with some ability to hydrogen bond. - Since compound
A is soluble in both water and ethanol, it likely has some polar groups that can
interact with the polar solvents.
Step 2: Consider the melting point of compound A and its relationship to
purity. - The melting point of a compound is affected by impurities: the presence
of impurities lowers the melting point. - The purer the compound, the higher
the melting point. - Since the melting point of compound A is 85
°
C, the student
can infer that the compound is relatively pure.
Step 3: Decide which solvent, water or ethanol, would be more suitable for
recrystallizing compound A. - Since compound A is soluble in both water and
ethanol, the choice between the two solvents comes down to maximizing purity.
- Water is a more polar solvent than ethanol and would dissolve a larger amount
of impurities due to its ability to hydrogen bond. - Ethanol, being a slightly
less polar solvent, would be a better choice for recrystallizing compound A to
obtain the purest crystals. - Therefore, the student should use ethanol as the
solvent for the recrystallization procedure of compound A.
23
Question 31
Question
A student is performing an experiment in the lab and needs to determine which
of the following compounds will be most soluble in water: hexane, ethanol, or
acetic acid. Provide a rationale for your answer based on the polarity of the
compounds.
Solution
In organic chemistry, solubility can be predicted based on the polarity of the
compounds involved. Polar compounds tend to dissolve in polar solvents like
water, while nonpolar compounds tend to dissolve in nonpolar solvents. Let’s
analyze the compounds given:
Hexane is a nonpolar molecule composed of carbon and hydrogen atoms
only. It is not soluble in water because of the significant difference in polarity
between hexane and water.
Ethanol is a polar molecule due to the presence of the hydroxyl (-OH)
group, which imparts polarity to the molecule. Ethanol can form hydrogen
bonds with water molecules, making it soluble in water to a moderate extent.
Acetic acid is also polar due to the presence of the carbonyl group and
the carboxyl group, which can both engage in hydrogen bonding. Acetic acid
is more polar than ethanol and can form stronger hydrogen bonds with water
molecules, making it highly soluble in water.
Therefore, among the compounds hexane, ethanol, and acetic acid, acetic
acid will be most soluble in water due to its higher polarity and ability to form
hydrogen bonds with water molecules.
Question 32
Question
An unknown compound X has a molecular formula of C6H14O. When 1.0 g
of compound X is dissolved in 10.0 g of water at 25
°
C, it forms a saturated
solution. Assuming the density of water is 1.0 g/mL, determine the solubility of
compound X in water at 25
°
C in g/mL. Additionally, predict whether compound
X is polar or nonpolar based on its solubility in water.
Solution
Step 1: Calculate the molarity of the saturated solution of compound X in
water.
Given: Mass of compound X = 1.0 g Volume of water = 10.0 g (since the density
of water is 1.0 g/mL, the volume of water in mL is also 10.0 mL) Molar mass
of compound X = 6(12.01) + 14(1.01) + 16.00 = 102.18 g/mol
24
Step 2: Calculate the molar solubility of compound X in water.
The molar solubility can be calculated using the formula:
Molarity = moles of solute
volume of solution in liters
First, calculate the moles of compound X:
Moles of C6H14O = mass
molar mass =1.0 g
102.18 g/mol = 0.0098 mol
Step 3: Calculate the solubility of compound X in water at 25
°
C.
The solubility can be calculated by dividing the mass of compound X by the
volume of water used:
Solubility = mass of solute
volume of solution =1.0 g
10.0 mL = 0.1 g/mL
Step 4: Predict whether compound X is polar or nonpolar based on its
solubility in water.
Since compound X is soluble in water, it is likely to be polar. Water is a
polar solvent, so polar compounds tend to be soluble in it due to the similar
intermolecular forces between the solute and the solvent.
Question 33
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C given
that the solubility product constant (Ksp) is 6.7×10−5mol2L−2.
Solution
Step 1: Write the equilibrium expression for the dissolution of benzoic acid.
The equation for the dissolution of benzoic acid in water is:
C6H5COOH ⇌C6H5COO−+ H+
The solubility of benzoic acid in water is equal to the concentration of
C6H5COO−ions.
Step 2: Write the expression for the solubility product constant (Ksp). The
Ksp expression is given by:
Ksp = [C6H5COO−][H+]
Given that benzoic acid is a weak acid, we can make an assumption that the
concentration of H+ions equals the concentration of C6H5COO−ions (H+=
C6H5COO−).
25
Step 3: Substitute the given Ksp value into the equilibrium expression. Sub-
stitute the given Ksp value and the assumption H+= C6H5COO−into the
equilibrium expression:
Ksp =x×x=x2
where xis the solubility of benzoic acid in water (in mol/L).
Step 4: Solve for the solubility of benzoic acid. From the equilibrium ex-
pression, we have:
x2= 6.7×10−5
x=p6.7×10−5
x≈0.0082 mol/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0082 mol/L.
Question 34
Question
An organic compound has the following molecular structure:
CH3−CH2−CH2−CH2−OH
Predict whether this compound would be soluble in water and explain your
reasoning.
Solution
To determine the solubility of the organic compound in water, we need to con-
sider its polarity and the polarity of water.
Step 1: Determine the polarity of the compound The molecular
structure contains a hydrocarbon chain and a hydroxyl group (-OH). The hy-
drocarbon chain is non-polar, while the hydroxyl group is polar due to the
presence of electronegative oxygen.
Step 2: Consider the polarity of water Water is a polar molecule be-
cause of the oxygen atom’s high electronegativity, causing an uneven distribution
of electrons.
Step 3: Predict solubility based on polarity Since ”like dissolves like,”
polar compounds are typically soluble in polar solvents, while non-polar com-
pounds are soluble in non-polar solvents. The presence of the polar hydroxyl
group in the compound suggests that it is polar. Therefore, the compound is
likely to be soluble in water due to the ability of polar substances to dissolve in
polar solvents.
Step 4: Conclusion Based on the analysis of the compound’s structure
and the polarity of water, we predict that the given organic compound (CH3−
CH2−CH2−CH2−OH)wouldbesolubleinwater.
26
Question 35
Question
A student wants to dissolve a compound, X, in water, but notices that it does not
dissolve. The student then tries to dissolve the compound in dichloromethane.
After several attempts, it is found that 1.5 g of the compound dissolves in 10 mL
of dichloromethane at room temperature. Calculate the solubility of compound
X in water in g/L at the same temperature. The density of dichloromethane at
room temperature is 1.33 g/mL.
Solution
Step 1: Calculate the molarity of compound X in dichloromethane. Given: Mass
of compound X = 1.5 g Volume of dichloromethane = 10 mL = 0.01 L Density
of dichloromethane = 1.33 g/mL
First, calculate the moles of compound X:
Moles of X = Mass of X
Molar mass of X
Step 2: Calculate the solubility of compound X in water. To calculate the
solubility of compound X in water, we need to consider the difference in polarity
between water and dichloromethane.
Since compound X is more soluble in dichloromethane (a nonpolar solvent)
than in water, it is likely that compound X is nonpolar. Therefore, it is safe
to assume that the compound cannot hydrogen bond with water, making its
solubility in water very low.
As a result, the solubility of compound X in water at room temperature is
considered negligible. So, the solubility of compound X in water at the given
temperature is approximately 0 g/L.
27
Question 2
Question
Determine which of the following pairs of compounds would be soluble in water,
and which would be more polar:
Compound A: Cyclohexanol Compound B: Octane
Solution
Step 1: Solubility in Water - Cyclohexanol has a hydroxyl group (−OH) which
can participate in hydrogen bonding with water molecules, making it soluble in
water. - Octane is a nonpolar hydrocarbon which does not have any functional
groups that can form significant intermolecular interactions with water, making
it insoluble in water.
Step 2: Polarity - Cyclohexanol is polar due to the presence of the hydroxyl
group, which can participate in hydrogen bonding. - Octane is nonpolar because
it consists of only carbon and hydrogen atoms and lacks any polar functional
groups.
Therefore, Compound A (Cyclohexanol) would be soluble in water and more
polar than Compound B (Octane).
Question 3
Question
Calculate the solubility (in g/L) of benzoic acid in water at 25
°
C. The solubility
of benzoic acid in water is 3.5 g/100 mL at that temperature.
Solution
Step 1: Start by converting the given solubility from g/100 mL to g/L.
Solubility in g/L = 3.5 g
100 mL×1000 mL
1 L
= 35 g/L
Step 2: Use the solubility of benzoic acid in water to perform the final
calculation.
Solubility in water = 35 g/L
Therefore, the solubility of benzoic acid in water at 25
°
C is 35 g/L.
2
Question 4
Question
Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. Given that
the solubility product constant, Ksp, for benzoic acid is 1.7×10−3at this
temperature.
Solution
Step 1: Write the dissociation equation for benzoic acid dissolving in water:
C7H6O2(s) ⇌C7H6O2(aq)
The solubility equilibrium expression can be given as:
Ksp = [C7H6O2]2
Step 2: Let x be the molar solubility of benzoic acid. Therefore, the equi-
librium concentrations are:
[C7H6O2] = x
Substitute the equilibrium concentrations into the solubility product con-
stant expression:
1.7×10−3= (x)2
x=p1.7×10−3
x= 0.0412 M
Therefore, the solubility of benzoic acid in water at 25◦C is 0.0412 M.
Question 5
Question
A student is given two organic compounds, compound A and compound B, and
is asked to determine which one is more soluble in water. Compound A has
a molecular weight of 150 g/mol and a log P value of 2.3, while compound B
has a molecular weight of 250 g/mol and a log P value of 1.8. Based on these
properties, which compound is expected to be more soluble in water? Justify
your answer.
Solution
To determine which compound is more soluble in water, we can use the par-
tition coefficient (P = solubility in octanol
solubility in water ) as a measure of how hydrophilic or
hydrophobic a compound is.
3
Step 1: Calculate the partition coefficient for compound A. The
partition coefficient (P) can be calculated using the formula:
P = 10log P
For compound A:
PA= 102.3≈199.53
Step 2: Calculate the partition coefficient for compound B. For
compound B:
PB= 101.8≈63.10
Step 3: Determine the solubility in water. Since a higher partition
coefficient indicates greater hydrophobicity, compound B is expected to be less
soluble in water compared to compound A. Therefore, compound A is expected
to be more soluble in water despite its lower molecular weight.
Question 6
Question
Calculate the solubility of compound X in water at 25
°
C. The partition coeffi-
cient of X between water and an organic solvent is 0.15. Assume the volume
change upon dissolution is negligible.
Solution
Step 1: Write the partition coefficient equation Step 2: Write the equation for
solubility of X in water Step 3: Calculate the solubility of X in water
Step 1: The partition coefficient (KOW) is defined as the ratio of solute
concentration in an organic solvent to the solute concentration in water. Math-
ematically,
KOW =[X in organic solvent]
[X in water] = 0.15
Step 2: Let’s assume the initial amount of compound X is x. If the solubility
of X in water is smol/L, then the solubility in the organic solvent will be x−s
mol/L. Using the partition coefficient equation, we have
0.15 = x−s
s
Step 3: Solving for s,
0.15s=x−s
0.15s+s=x
s(0.15 + 1) = x
s=x
1.15
4
Since the volume change upon dissolution is negligible, the moles of X before
and after dissolution are the same. Therefore, x= solubility of X in water at
25
°
C.
Thus, the solubility of compound X in water at 25
°
C is x
1.15 =solubility of X in organic solvent
1.15 .
Question 7
Question
Calculate the solubility of benzoic acid in water at 25◦C. The solubility of ben-
zoic acid in water is 3.4 g/L at 100◦C and its partition coefficient between water
and benzene at room temperature is 10.5. Assume the density of water is 1.0
g/mL and benzene is 0.88 g/mL at room temperature.
Solution
Step 1: Calculate the partition coefficient Kat 25◦C using the given value at
another temperature. Given that T1= 100◦C, T2= 25◦C, K1= 10.5, and
∆Hsol =−33 kJ/mol, we can use the van’t Hoff equation:
ln K2
K1=−∆Hsol
R1
T2
−1
T1
where R= 8.314 J/mol ·K=0.008314 kJ/mol ·K.
ln K
10.5=−
−33
0.008314 1
298 −1
373
ln K
10.5= 14.2581
K
10.5=e14.2581
K= 10.5×e14.2581
K≈2.04 ×106
Step 2: Calculate the molar solubility of benzoic acid in water. Given that
K= 2.04 ×106, let the molar solubility of benzoic acid in water at 25◦C be x
mol/L. The equilibrium between the solid benzoic acid and the dissolved benzoic
acid is represented by:
C6H5COOH(s)⇌C6H5COOH(aq)
The equilibrium constant Keq is given by:
Keq =[C6H5COOH(aq)]
[C6H5COOH(s)] ≈x
3.4= 2.04 ×106
5
Solving for x:
x= 2.04 ×106×3.4
x≈6.94 ×106mol/L
Step 3: Calculate the mass solubility of benzoic acid in water. Given that the
molar mass of benzoic acid is 122.12 g/mol, we can convert the molar solubility
to mass solubility:
Mass solubility = 6.94 ×106mol/L ×122.12 g/mol
Mass solubility ≈8.47 ×105g/L
Thus, the solubility of benzoic acid in water at 25◦C is approximately 847 g/L.
Question 8
Question
A student is trying to determine the solubility of compound X in different sol-
vents. Compound X is known to be nonpolar and has a molecular weight of 150
g/mol. The student finds that compound X is soluble in hexane but insoluble
in water. Based on this information, calculate the solubility (in grams per liter)
of compound X in hexane.
(Hint: Use the molecular weight of the compound to calculate the molarity
in hexane, and then convert this to solubility in grams per liter.)
Solution
Step 1: Calculate the molarity of compound X in hexane
The molarity of compound X in hexane can be calculated as:
Molarity = grams of solute
molecular weight of solute ×volume of solvent (L)
Since the molecular weight of compound X is 150 g/mol, we can use this to
calculate the molarity in hexane.
Step 2: Convert the molarity to solubility in grams per liter
Given that the student found compound X to be soluble in hexane, the
calculated molarity can be converted to solubility in grams per liter using the
relationship:
Solubility (g/L) = Molarity ×Molecular weight of solute
6
Question 9
Question
A student is conducting a solubility experiment in which they dissolve 0.050
moles of a nonpolar organic compound in 50 mL of water at 25
°
C. The student
notices that the compound partially dissolves. Calculate the solubility of the
compound in water in mol/L. Assume the volume of the organic compound is
negligible.
Solution
Step 1: Calculate the molarity of the water solution. Given: - Moles of organic
compound = 0.050 mol - Volume of water = 50 mL = 0.050 L
The molarity of the water solution is calculated using the formula:
Molarity =Number of moles
Volume in liters
Plugging in the values:
Molarity =0.050 mol
0.050 L
Molarity = 1.0 mol/L
Step 2: Calculate the solubility of the compound in water. Since the com-
pound partially dissolves, let’s assume x moles of the compound dissolves in 1
L water. The equilibrium expression for the dissolution of the compound is:
Compound ⇌(aq)
The solubility product expression is:
Ksp = [Compound] = x
The solubility product constant (Ksp) for a nonpolar compound is assumed
to be very small. Since the compound partially dissolves, we can assume that
the equilibrium concentration of the compound is much less than 1.0 mol/L.
Therefore, the solubility of the compound in water at 25
°
C is very low, close
to or less than 1.0 mol/L.
Question 10
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The Ksp
of benzoic acid in water is 8.3×10−5. Assume the density of water is 1.0 g/mL.
7
Solution
Step 1: Write the dissociation reaction of benzoic acid in water:
C6H5COOH ⇌C6H5COO−+ H+
Step 2: Construct the equilibrium expression for the dissociation of benzoic
acid:
Ksp = [C6H5COO−][H+]
Step 3: Let x be the molar solubility of benzoic acid. Since benzoic acid is a
weak acid, we can approximate the equilibrium concentrations of C6H5COO−
and H+as x. Therefore, the expression becomes:
Ksp = (x)(x)
Ksp =x2
Step 4: Substitute the given Ksp value into the equilibrium expression and
solve for x:
8.3×10−5=x2
x=p8.3×10−5
x≈0.0091 M
Step 5: Calculate the solubility of benzoic acid in g/L:
Molar mass of benzoic acid = (6×12.01)+(5×1.01)+12.01+16.00+1.01 = 122.12 g/mol
Solubility in g/L = (0.0091 mol/L) ×(122.12 g/mol) ×(1 L/1000 mL)
Solubility in g/L ≈0.0111 g/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0111 g/L.
Question 11
Question
A student needs to determine the solubility of a compound in water at room
temperature. The student knows that the compound is a carboxylic acid and
has a molecular weight of 122 g/mol. The student performed a solubility test
and found that 0.5 g of the compound dissolves in 10 mL of water. Calculate
the solubility of the compound in mol/L.
8
Solution
Step 1: Calculate the molar mass of the compound. Step 2: Determine the
number of moles of the compound that dissolve in 10 mL of water. Step 3:
Convert the volume of water to liters. Step 4: Calculate the solubility of the
compound in mol/L.
Step 1: The molar mass of the compound is 122 g/mol.
Step 2: The number of moles of the compound that dissolve in 10 mL of
water can be calculated using the given mass and molar mass. Number of moles
= Mass / Molar mass Number of moles = 0.5 g / 122 g/mol
Step 3: Convert the volume of water from milliliters to liters. 1 mL = 0.001
L Volume of water = 10 mL * 0.001 L/mL
Step 4: Now, calculate the solubility of the compound in mol/L. Solubility
(in mol/L) = Number of moles / Volume of water
Therefore, the solubility of the compound in water at room temperature is
0.5 g/122 g/mol
10 mL ×0.001 L/mL
mol/L.
Question 12
Question
A student is given a sample of compound X and is asked to determine its solu-
bility in water. After running solubility tests, the student finds that compound
X is soluble in both water and hexane. Based on this information, determine
the possible functional groups present in compound X and explain the reasoning
behind these possibilities.
Solution
Step 1: Since compound X is soluble in water, it must possess some polar func-
tional groups that can interact with water molecules through hydrogen bonding
or dipole-dipole interactions.
Step 2: The solubility of compound X in hexane suggests the presence of
nonpolar functional groups that can interact with nonpolar solvents like hexane
through London dispersion forces.
Step 3: Keeping in mind the solubility information in both water and hexane,
the possible functional groups in compound X could include alcohols, amines,
carboxylic acids, and esters.
Step 4: Alcohols have a polar hydroxyl (-OH) functional group that allows
them to form hydrogen bonds with water molecules, making them soluble in
water. However, the nonpolar alkyl group in alcohols also allows them to dissolve
in nonpolar solvents like hexane.
9
Step 5: Amines contain a polar amino (-NH2) functional group, which can
form hydrogen bonds with water molecules, leading to their solubility in water.
The alkyl group in amines contributes to their solubility in nonpolar solvents
like hexane.
Step 6: Carboxylic acids have a polar carboxyl (-COOH) group that allows
them to form hydrogen bonds with water, making them soluble in water. The
nonpolar alkyl group in carboxylic acids enables them to dissolve in nonpolar
solvents like hexane.
Step 7: Esters contain a polar carbonyl (C=O) group and can form hydrogen
bonds with water molecules, leading to their solubility in water. The nonpolar
alkyl groups in esters contribute to their solubility in nonpolar solvents like
hexane.
Step 8: Therefore, compound X could potentially contain alcohols, amines,
carboxylic acids, or esters as functional groups based on its solubility charac-
teristics in both water and hexane.
Question 13
Question
A student is given a white solid compound to identify in the laboratory. The
compound is sparingly soluble in water but dissolves readily in ether. Upon
adding HCl to the white solid, a white precipitate is formed. On the basis of
the solubility properties described, which functional group or groups are most
likely present in the compound?
Solution
Step 1: The compound is sparingly soluble in water but dissolves readily in
ether. This information suggests that the compound is nonpolar or has a low
polarity.
Step 2: The compound dissolves in ether, a nonpolar solvent, indicating that
the compound is likely nonpolar or has a low polarity. Nonpolar compounds
tend to dissolve well in nonpolar solvents such as ether.
Step 3: Upon addition of HCl, a white precipitate is formed. This suggests
the presence of a functional group that can react with HCl to form an insoluble
product.
Step 4: The functional group that reacts with HCl to form an insoluble
product is likely an amine group (RNH2, R2NH, or R3N). Amines react with
HCl to form ammonium chloride salts, which are typically insoluble in water
and precipitate out.
Step 5: Therefore, the functional group most likely present in the compound
based on the given solubility properties is an amine group.
10
Question 14
Question
An organic compound, X, has a molecular weight of 86 g/mol. When 0.5 g of
compound X is dissolved in 10 mL of water at 25
°
C, it forms a clear solution.
However, when 0.5 g of compound X is dissolved in 10 mL of diethyl ether at
25
°
C, it forms a cloudy solution. Determine whether compound X is more likely
to be polar or nonpolar.
Solution
Step 1: Calculate the molarity of compound X in water solution. Given: - Mass
of X = 0.5 g - Volume of water = 10 mL = 0.01 L - Molecular weight of X =
86 g/mol
First, calculate the number of moles of X:
Moles of X = Mass of X
Molecular weight of X =0.5 g
86 g/mol
Moles of X ≈0.00581 mol
Next, calculate the molarity of X in water:
Molarity = Moles of X
Volume of solution =0.00581 mol
0.01 L
Molarity = 0.581 M
Step 2: Calculate the molarity of compound X in diethyl ether solution.
Given: - Volume of diethyl ether = 10 mL = 0.01 L
Since the compound X forms a cloudy solution in diethyl ether, it is less
soluble in diethyl ether than in water. This indicates that compound X is more
likely to be polar due to the polarity of water.
Thus, compound X is more likely to be a polar compound.
Question 15
Question
Calculate the solubility of compound X in water at 25
°
C. The observed solubility
is 0.04 mol/L. The partition coefficient of X between water and benzene is 10.
Solution
Step 1: Calculate the solubility of compound X in benzene. Given: Partition
coefficient (Kow) = 10 Let the solubility of X in benzene be xmol/L.
11
The partition coefficient is defined as:
Kow =Concentration of X in benzene
Concentration of X in water
10 = x
0.04
x= 0.4 mol/L
Therefore, the solubility of compound X in benzene is 0.4 mol/L.
Step 2: Calculate the solubility of X in water. Let the solubility of X in
water be ymol/L.
The observation is that the solubility of X in water is 0.04 mol/L, and the
solubility in benzene is 0.4 mol/L. This means the solute prefers to dissolve in
benzene over water.
Hence, y= 0.04 mol/L.
Therefore, the solubility of compound X in water at 25
°
C is 0.04 mol/L.
Question 16
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The Ksp
for benzoic acid at 25
°
C is 1.3×10−3.
Solution
Step 1: Write the dissociation equilibrium equation for benzoic acid:
C6H5COOH ⇌C6H5COO−+ H+
Step 2: Write the expression for the Ksp of benzoic acid:
Ksp = [C6H5COO−][H+]
Step 3: Let x be the solubility of benzoic acid. Since benzoic acid is a weak
acid, its solubility is equal to its dissociation in water as per the equation in
Step 1:
C6H5COOH(s)⇌C6H5COO−(aq)+H+(aq)
The equilibrium concentrations will be: [C6H5COO−] = xand [H+] = x
Step 4: Substitute the equilibrium concentrations into the Ksp expression:
Ksp =x×x=x2
Step 5: Plug in the given Ksp value:
1.3×10−3=x2
12
Step 6: Solve for x:
x=p1.3×10−3
x≈0.0361 M
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0361 M.
Question 17
Question
A student is conducting an experiment in which they need to dissolve 0.10 moles
of compound X in 500 mL of water. Compound X has a molar mass of 120 g/mol
and a solubility in water of 0.5 g/L. Is it possible to dissolve all 0.10 moles of
compound X in 500 mL of water? If not, calculate the maximum amount of
compound X that can be dissolved in the water.
Solution
Step 1: Calculate the maximum amount of compound X in grams that can be
dissolved in 500 mL of water. Given: - Molar mass of compound X = 120 g/mol
- Solubility of compound X in water = 0.5 g/L - Number of moles of compound
X to be dissolved = 0.10 moles - Volume of water available = 500 mL
First, convert 500 mL to liters:
500 mL = 500 ×10−3L=0.500 L
To find the maximum amount of compound X that can be dissolved in water,
we can use the solubility information:
Maximum amount of compound X = Solubility ×Volume of water
Maximum amount of compound X = 0.5 g/L ×0.500 L
Maximum amount of compound X = 0.25 g
Step 2: Calculate the amount of compound X in grams that corresponds to
0.10 moles. Given: - Number of moles of compound X to be dissolved = 0.10
moles
To find the amount of compound X in grams corresponding to 0.10 moles,
we can use the molar mass:
Amount of compound X for 0.10 moles = Number of moles ×Molar mass
Amount of compound X for 0.10 moles = 0.10 moles ×120 g/mol
Amount of compound X for 0.10 moles = 12 g
Since the amount of compound X for 0.10 moles (12 g) is greater than the
maximum amount that can be dissolved (0.25 g), it is not possible to dissolve
all 0.10 moles of compound X in 500 mL of water. The maximum amount that
can be dissolved is 0.25 g.
13
Question 18
Question
A student is given two compounds, compound A and compound B, and is asked
to predict which compound is more soluble in water based on their structures.
Compound A has a molecular weight of 150 g/mol and consists of a long carbon
chain with a hydroxyl group (-OH) at one end. Compound B has a molecular
weight of 120 g/mol and consists of a cyclic structure with a carbonyl group
(C=O). Calculate the HLB (hydrophilic-lipophilic balance) value for each com-
pound and determine which compound is more likely to be soluble in water.
Solution
Step 1: Calculate the HLB value for compound A.
HLBA = 20 ×nOH
MW + 16 ×nCOOH
MW + 1.8×nester
MW
Given that compound A has a molecular weight of 150 g/mol and contains
one hydroxyl group (-OH):
HLBA = 20 ×1
150=20
150 = 0.133
Step 2: Calculate the HLB value for compound B.
HLBB = 20 ×nOH
MW + 16 ×nCOOH
MW + 1.8×nester
MW
Given that compound B has a molecular weight of 120 g/mol and contains
one carbonyl group (C=O):
HLBB = 16 ×1
120=16
120 = 0.133
Step 3: Compare the HLB values for compound A and compound B.
Since both compound A and compound B have the same HLB value of
0.133, this means that they both have similar hydrophilic-lipophilic balance
values. Therefore, based solely on the HLB values, it is difficult to predict
which compound is more likely to be soluble in water. Other factors such as the
specific interactions between the compounds and water molecules would need
to be considered to determine solubility.
Question 19
Question
An organic compound X has a molecular weight of 150 g/mol and an experi-
mentally determined solubility of 5 g/L at 25
°
C. Determine the solubility of X
in hexane (C6H14) at the same temperature. The density of hexane is 0.659
g/mL.
14
Solution
Step 1: Calculate the molar solubility of compound X in water. Given that the
solubility of X in water is 5 g/L and its molecular weight is 150 g/mol, we can
calculate the molar solubility in water as follows:
Molar solubility in water = 5 g/L
150 g/mol =1
30 mol/L
Step 2: Calculate the partition coefficient, P, of compound X between water
and hexane. The partition coefficient is defined as the ratio of the solubility of
a compound in two immiscible solvents.
P=Solubility of X in hexane
Solubility of X in water
Step 3: Using the partition coefficient to find the solubility of X in hexane.
We know that the solubility of X in water is 5 g/L and the molar solubility in
water is 1
30 mol/L. Let the solubility of X in hexane be xg/L. From the partition
coefficient formula, we have:
P=x
5=1
30
Solving for x, we get:
x=5
30 =1
6g/L
Therefore, the solubility of compound X in hexane at 25
°
C is 1
6g/L.
Question 20
Question
A student wants to separate a mixture of benzoic acid (solubility in water at
20
°
C is 3.5 g/L) and biphenyl (solubility in water at 20
°
C is 0.02 g/L) using the
principle of solubility. The student knows that the two compounds have different
solubilities in water due to the difference in polarity. Discuss the solubility of
benzoic acid and biphenyl in water based on their polarity.
Solution
Step 1: Benzoic acid is a polar molecule due to the presence of a carboxylic
acid group, which allows it to form hydrogen bonds with water molecules. This
results in higher solubility in water compared to nonpolar biphenyl.
Step 2: Biphenyl is a nonpolar molecule because it consists of two benzene
rings connected by a single bond. The lack of polar functional groups makes
biphenyl insoluble or poorly soluble in water.
Step 3: In the separation process, the student can exploit the difference in
solubility to dissolve benzoic acid in water and leave biphenyl undissolved. This
15
can be achieved by selecting an appropriate solvent that will dissolve benzoic
acid but not biphenyl, allowing for the separation of the two compounds based
on their polarity.
Step 4: By understanding the solubility and polarity of organic compounds,
the student can effectively design a separation strategy for mixtures containing
both polar and nonpolar compounds.
Question 21
Question
A student is conducting an experiment in the laboratory and produces a solution
by mixing 50.0 mL of water and 50.0 mL of diethyl ether. The student wants
to know if the resulting solution is homogeneous or heterogeneous based on its
solubility and polarity. Given that water has a density of 1.00 g/mL and diethyl
ether has a density of 0.71 g/mL, calculate the solubility parameter () for water
and diethyl ether. Based on the solubility parameters, determine if the solution
is likely to be homogeneous or heterogeneous.
Solution
Step 1: Calculate the solubility parameter () for water and diethyl ether using
the equation:
δ=qδ2
d+δ2
p
where δdis the dispersion component and δpis the polar component of the
solubility parameter.
For water:
δd,water = 14.6 MPa0.5
δp,water = 20.0 MPa0.5
δwater =p14.62+ 20.02= 25.0 MPa0.5
For diethyl ether:
δd,ether = 14.5 MPa0.5
δp,ether = 5.8 MPa0.5
δether =p14.52+ 5.82= 15.6 MPa0.5
Step 2: Compare the solubility parameters of water and diethyl ether. Since
the solubility parameters of water (25.0 MPa0.5) and diethyl ether (15.6 MPa0.5)
are significantly different, the solution is likely to be heterogeneous. Water,
being more polar than diethyl ether, will not dissolve diethyl ether completely,
resulting in a heterogeneous solution.
16
Question 22
Question
Indicate whether each of the following compounds is more soluble in water or
in hexane based on their polarity:
A) Butanol (CHOH)
B) Pentanoic acid (CHCOOH)
C) Toluene (CH)
D) Acetone (CHCOCH)
Solution
Step 1: Determine the polarity of each compound based on its functional groups.
A) Butanol: contains a hydroxyl (-OH) group, making it a polar com-
pound.
B) Pentanoic acid: contains a carboxyl (-COOH) group, making it a polar
compound.
C) Toluene: a nonpolar compound as it consists of only carbon and hy-
drogen atoms.
D) Acetone: a polar compound due to the presence of the carbonyl group.
Step 2: Evaluate the solubility of each compound in water or hexane based
on their polarity.
A) Butanol: more soluble in water due to its polar nature and ability to
form hydrogen bonds with water molecules.
B) Pentanoic acid: soluble in water due to the polar carboxyl group that
can interact with water molecules through hydrogen bonding.
C) Toluene: more soluble in hexane, a nonpolar solvent, as both are non-
polar compounds that can interact through London dispersion forces.
D) Acetone: soluble in both water and hexane, but more soluble in water
due to its polar nature and ability to form hydrogen bonds with water
molecules.
Therefore, the solubility of the compounds in either water or hexane can
be predicted based on their polarity and ability to interact with the solvent
molecules.
17
Question 23
Question
Determine which compound, A or B, would be more soluble in water based on
their respective chemical structures and polarities. Justify your answer.
Compound A: 1-hexanol (CH3(CH2)5OH) Compound B: 1-hexanone (CH3(CH2)4COCH3)
Solution
Step 1: Analyze the chemical structures of the compounds.
Compound A (1-hexanol) contains a hydroxyl group (OH), making it a polar
molecule. The compound exhibits a polar covalent bond between the carbon
and oxygen atoms of the hydroxyl group.
Compound B (1-hexanone) contains a carbonyl group (CO), resulting in
a polar molecule due to the electronegativity difference between carbon and
oxygen in the carbonyl group.
Step 2: Determine the solubility of each compound in water based on polar-
ity.
Water is a polar molecule due to its bent structure and unequal sharing
of electrons between oxygen and hydrogen atoms. Polar solutes are generally
soluble in polar solvents like water.
Since both compounds A and B are polar molecules, both are expected to be
soluble in water due to the ability of polar solutes to dissolve in polar solvents.
However, the presence of an additional hydrogen bonding site in compound A
(OH group) compared to compound B (carbonyl group) may lead to stronger
interactions with water molecules, potentially making compound A more soluble
in water than compound B. Thus, compound A (1-hexanol) is expected to be
more soluble in water than compound B (1-hexanone).
Question 24
Question
An organic compound with the molecular formula C6H12Ois found to be soluble
in water. However, it is insoluble in hexane (C6H14), a nonpolar solvent. Pro-
vide a possible structure for this compound and explain its solubility behavior
in terms of its polarity.
Solution
Step 1: Calculate the degree of unsaturation (DU) to determine the compound’s
possible structures. The formula for degree of unsaturation is:
DU =2C+ 2 −H+X−N
2
18
Where: C= 6 (number of carbon atoms), H= 12 (number of hydrogen
atoms), and X= 16 (number of halogen atoms).
Since there are no halogens or nitrogen in the formula, we have:
DU =2∗6+2−12
2= 1
This indicates that the compound is likely to contain a double bond or a
ring.
Step 2: Given that the compound is soluble in water but insoluble in hexane,
it suggests that the compound is polar. Water is a polar solvent, while hexane
is nonpolar. A possible structure that fits the molecular formula C6H12Oand
is polar is a cyclic ether:
CH3−CH2−CH2−O−CH2−CH2−CH3
The presence of the oxygen atom in the ether functional group makes the
molecule polar. The oxygen atom is more electronegative than carbon and
hydrogen, creating a partial negative charge on the oxygen and partial positive
charges on the carbons. This polarity allows the compound to form hydrogen
bonds with water molecules, making it soluble in water. However, the nonpolar
hexane cannot form the necessary interactions to dissolve the compound.
Question 25
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C. The
given Ksp value for benzoic acid is 1.7×10−3mol/L.
Solution
Step 1: Write the dissociation equation for benzoic acid in water:
C6H5COOH ⇌C6H5COO−+H+
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [C6H5COO−][H+]
Step 3: Since benzoic acid is a weak acid, we need to assume that there
is negligible dissociation of C6H5COOH (essentially none). Therefore, we can
assume that the concentrations of C6H5COO−and H+are equal to the initial
solubility of benzoic acid. Let’s denote the solubility of benzoic acid as x.
Step 4: Substitute the expressions for the concentrations into the Ksp ex-
pression:
1.7×10−3=x·x
1.7×10−3=x2
19
Step 5: Solve for xto find the solubility of benzoic acid:
x=p1.7×10−3
x≈0.0412 mol/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0412 mol/L.
Question 26
Question
A student is performing a solubility test for a compound and observes that it
dissolves in ether but not in water. The student also determines the compound’s
partition coefficient between water and ether to be 3.2. Calculate the percentage
of the compound that would dissolve in ether if 10 g of the compound were added
to 100 mL of ether.
Solution
Step 1: Calculate the mass of the compound that would dissolve in water and in
ether. Let the mass of the compound that dissolves in water be xg. Therefore,
the mass of the compound that dissolves in ether would be 10 −xg.
Step 2: Use the definition of the partition coefficient to set up an equation.
Kp=[Compound in ether]
[Compound in water],
where Kp= 3.2.
Step 3: Express the concentrations in terms of masses and volumes. Given
that the compound is added to 100 mL of ether: - The concentration of the
compound in ether is 10 −x
100 g/mL. - The concentration of the compound in
water is x
100 g/mL.
Step 4: Substitute the concentrations into the partition coefficient equation.
3.2 = 10 −x
x.
Step 5: Solve the equation for x.
3.2x= 10 −x
4.2x= 10
x=10
4.2≈2.38 g
20
Thus, approximately 2.38 g of the compound dissolves in water, and 10 −
2.38 = 7.62 g of the compound dissolves in ether.
Step 6: Calculate the percentage of the compound that dissolves in ether.
Percentage in ether = Mass in ether
Total mass ×100
Percentage in ether = 7.62
10 ×100
Percentage in ether ≈76.2%
Therefore, approximately 76.2
Question 27
Question
Calculate the solubility of benzoic acid (C7H6O2) in water at 25
°
C. The solu-
bility product constant for benzoic acid in water is 6.2×10−3M2.
Solution
Step 1: Write the equilibrium equation representing the dissolution of benzoic
acid in water. Step 2: Write the expression for the solubility product constant
(Ksp). Step 3: Define the variables and set up the equilibrium expression for
benzoic acid. Step 4: Solve for the molar solubility of benzoic acid in water.
Question 28
Question
A compound has a solubility of 0.15 g/100 mL in water at 30
°
C. Calculate the
solubility product constant (Ksp) for this compound in water at 30
°
C. Assume
the compound completely dissociates in water.
Solution
Step 1: Calculate the molar solubility of the compound.
Molar solubility = Solubility (g/L)
Molar mass (g/mol)
Molar solubility = 0.15 g/100 mL ×10 mL/L
M
Molar solubility = 0.015 g/L
M
21
Step 2: Calculate the Ksp using the molar solubility. The compound com-
pletely dissociates into its ions. If the compound is represented as AxBy, the
dissociation equation would be AxBy⇌xAz++yBw−. Then, the Ksp expres-
sion is:
Ksp = [Az+]x[Bw−]y
Since there is complete dissociation, we can say that:
[Az+] = x×Molar solubility = x×0.015 g/L
M
[Bw−] = y×Molar solubility = y×0.015 g/L
M
Substitute these expressions into the Ksp expression:
Ksp =x×0.015 g/L
Mxy×0.015 g/L
My
Question 29
Question
Compound A has a solubility of 0.15 g/L in water at 25
°
C. Compound B has a
solubility of 0.30 g/L in water at 25
°
C. Which compound is more polar? Justify
your answer with calculations.
Solution
Step 1: Calculate the molar solubility of each compound. Let’s assume the
molar mass of compound A is MAg/mol and the molar mass of compound B
is MBg/mol.
The molar solubility of compound A (SA) can be calculated as:
SA=0.15 g/L
MAg/mol
The molar solubility of compound B (SB) can be calculated as:
SB=0.30 g/L
MBg/mol
Step 2: Compare the molar solubilities of compounds A and B. To compare
the polarities of compounds A and B, we can compare their molar solubilities.
Higher molar solubility indicates a more polar compound.
If SA> SB, then compound A is more polar. If SA< SB, then compound
B is more polar.
Step 3: Use experimental data to compare the polarities of compounds A
and B. Assume MA= 100 g/mol and MB= 150 g/mol for the purpose of
comparison.
22
Calculate the molar solubility of compound A:
SA=0.15 g/L
100 g/mol = 0.0015 mol/L
Calculate the molar solubility of compound B:
SB=0.30 g/L
150 g/mol = 0.0020 mol/L
Step 4: Compare the molar solubilities of compounds A and B: Since SB>
SA, compound B is more polar than compound A based on their experimental
solubilities.
Question 30
Question
A student is trying to decide which solvent to use in a recrystallization procedure
for a compound. Compound A has a melting point of 85
°
C and is known to
be soluble in both ethanol and water. The student knows that the solubility of
a compound is related to the polarity of the solvent, with more polar solvents
dissolving more polar compounds. The student also knows that the melting
point of a compound is related to its purity, with impure compounds having
lower melting points. Which solvent, water or ethanol, would be more suitable
for recrystallizing compound A to obtain the purest crystals? Justify your
answer.
Solution
Step 1: Determine the polarity of compound A in relation to water and ethanol.
- Water is a highly polar solvent due to its ability to hydrogen bond. - Ethanol
is also a polar solvent with some ability to hydrogen bond. - Since compound
A is soluble in both water and ethanol, it likely has some polar groups that can
interact with the polar solvents.
Step 2: Consider the melting point of compound A and its relationship to
purity. - The melting point of a compound is affected by impurities: the presence
of impurities lowers the melting point. - The purer the compound, the higher
the melting point. - Since the melting point of compound A is 85
°
C, the student
can infer that the compound is relatively pure.
Step 3: Decide which solvent, water or ethanol, would be more suitable for
recrystallizing compound A. - Since compound A is soluble in both water and
ethanol, the choice between the two solvents comes down to maximizing purity.
- Water is a more polar solvent than ethanol and would dissolve a larger amount
of impurities due to its ability to hydrogen bond. - Ethanol, being a slightly
less polar solvent, would be a better choice for recrystallizing compound A to
obtain the purest crystals. - Therefore, the student should use ethanol as the
solvent for the recrystallization procedure of compound A.
23
Question 31
Question
A student is performing an experiment in the lab and needs to determine which
of the following compounds will be most soluble in water: hexane, ethanol, or
acetic acid. Provide a rationale for your answer based on the polarity of the
compounds.
Solution
In organic chemistry, solubility can be predicted based on the polarity of the
compounds involved. Polar compounds tend to dissolve in polar solvents like
water, while nonpolar compounds tend to dissolve in nonpolar solvents. Let’s
analyze the compounds given:
Hexane is a nonpolar molecule composed of carbon and hydrogen atoms
only. It is not soluble in water because of the significant difference in polarity
between hexane and water.
Ethanol is a polar molecule due to the presence of the hydroxyl (-OH)
group, which imparts polarity to the molecule. Ethanol can form hydrogen
bonds with water molecules, making it soluble in water to a moderate extent.
Acetic acid is also polar due to the presence of the carbonyl group and
the carboxyl group, which can both engage in hydrogen bonding. Acetic acid
is more polar than ethanol and can form stronger hydrogen bonds with water
molecules, making it highly soluble in water.
Therefore, among the compounds hexane, ethanol, and acetic acid, acetic
acid will be most soluble in water due to its higher polarity and ability to form
hydrogen bonds with water molecules.
Question 32
Question
An unknown compound X has a molecular formula of C6H14O. When 1.0 g
of compound X is dissolved in 10.0 g of water at 25
°
C, it forms a saturated
solution. Assuming the density of water is 1.0 g/mL, determine the solubility of
compound X in water at 25
°
C in g/mL. Additionally, predict whether compound
X is polar or nonpolar based on its solubility in water.
Solution
Step 1: Calculate the molarity of the saturated solution of compound X in
water.
Given: Mass of compound X = 1.0 g Volume of water = 10.0 g (since the density
of water is 1.0 g/mL, the volume of water in mL is also 10.0 mL) Molar mass
of compound X = 6(12.01) + 14(1.01) + 16.00 = 102.18 g/mol
24
Step 2: Calculate the molar solubility of compound X in water.
The molar solubility can be calculated using the formula:
Molarity = moles of solute
volume of solution in liters
First, calculate the moles of compound X:
Moles of C6H14O = mass
molar mass =1.0 g
102.18 g/mol = 0.0098 mol
Step 3: Calculate the solubility of compound X in water at 25
°
C.
The solubility can be calculated by dividing the mass of compound X by the
volume of water used:
Solubility = mass of solute
volume of solution =1.0 g
10.0 mL = 0.1 g/mL
Step 4: Predict whether compound X is polar or nonpolar based on its
solubility in water.
Since compound X is soluble in water, it is likely to be polar. Water is a
polar solvent, so polar compounds tend to be soluble in it due to the similar
intermolecular forces between the solute and the solvent.
Question 33
Question
Calculate the solubility of benzoic acid (C6H5COOH) in water at 25
°
C given
that the solubility product constant (Ksp) is 6.7×10−5mol2L−2.
Solution
Step 1: Write the equilibrium expression for the dissolution of benzoic acid.
The equation for the dissolution of benzoic acid in water is:
C6H5COOH ⇌C6H5COO−+ H+
The solubility of benzoic acid in water is equal to the concentration of
C6H5COO−ions.
Step 2: Write the expression for the solubility product constant (Ksp). The
Ksp expression is given by:
Ksp = [C6H5COO−][H+]
Given that benzoic acid is a weak acid, we can make an assumption that the
concentration of H+ions equals the concentration of C6H5COO−ions (H+=
C6H5COO−).
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Step 3: Substitute the given Ksp value into the equilibrium expression. Sub-
stitute the given Ksp value and the assumption H+= C6H5COO−into the
equilibrium expression:
Ksp =x×x=x2
where xis the solubility of benzoic acid in water (in mol/L).
Step 4: Solve for the solubility of benzoic acid. From the equilibrium ex-
pression, we have:
x2= 6.7×10−5
x=p6.7×10−5
x≈0.0082 mol/L
Therefore, the solubility of benzoic acid in water at 25
°
C is approximately
0.0082 mol/L.
Question 34
Question
An organic compound has the following molecular structure:
CH3−CH2−CH2−CH2−OH
Predict whether this compound would be soluble in water and explain your
reasoning.
Solution
To determine the solubility of the organic compound in water, we need to con-
sider its polarity and the polarity of water.
Step 1: Determine the polarity of the compound The molecular
structure contains a hydrocarbon chain and a hydroxyl group (-OH). The hy-
drocarbon chain is non-polar, while the hydroxyl group is polar due to the
presence of electronegative oxygen.
Step 2: Consider the polarity of water Water is a polar molecule be-
cause of the oxygen atom’s high electronegativity, causing an uneven distribution
of electrons.
Step 3: Predict solubility based on polarity Since ”like dissolves like,”
polar compounds are typically soluble in polar solvents, while non-polar com-
pounds are soluble in non-polar solvents. The presence of the polar hydroxyl
group in the compound suggests that it is polar. Therefore, the compound is
likely to be soluble in water due to the ability of polar substances to dissolve in
polar solvents.
Step 4: Conclusion Based on the analysis of the compound’s structure
and the polarity of water, we predict that the given organic compound (CH3−
CH2−CH2−CH2−OH)wouldbesolubleinwater.
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Question 35
Question
A student wants to dissolve a compound, X, in water, but notices that it does not
dissolve. The student then tries to dissolve the compound in dichloromethane.
After several attempts, it is found that 1.5 g of the compound dissolves in 10 mL
of dichloromethane at room temperature. Calculate the solubility of compound
X in water in g/L at the same temperature. The density of dichloromethane at
room temperature is 1.33 g/mL.
Solution
Step 1: Calculate the molarity of compound X in dichloromethane. Given: Mass
of compound X = 1.5 g Volume of dichloromethane = 10 mL = 0.01 L Density
of dichloromethane = 1.33 g/mL
First, calculate the moles of compound X:
Moles of X = Mass of X
Molar mass of X
Step 2: Calculate the solubility of compound X in water. To calculate the
solubility of compound X in water, we need to consider the difference in polarity
between water and dichloromethane.
Since compound X is more soluble in dichloromethane (a nonpolar solvent)
than in water, it is likely that compound X is nonpolar. Therefore, it is safe
to assume that the compound cannot hydrogen bond with water, making its
solubility in water very low.
As a result, the solubility of compound X in water at room temperature is
considered negligible. So, the solubility of compound X in water at the given
temperature is approximately 0 g/L.
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