CHEM 122 - GENERAL CHEMISTRY
II - Titration curves and calculations
Question Bank - Set 9
Liberty University
Question 1
Question
A 25.0 mL sample of 0.100 M ethanoic acid (CH3COOH) is titrated with 0.150
M sodium hydroxide (NaOH). Calculate the pH at the following volumes of
added base:
1. 0.00 mL
2. 12.5 mL
3. 25.0 mL
4. 35.0 mL
5. 50.0 mL
Given that the Kaof ethanoic acid is 1.8×10−5.
Solution
Step 1: Calculate the initial moles of ethanoic acid present. The initial moles
of ethanoic acid can be calculated using the formula:
moles = concentration ×volume (L)
For the initial 25.0 mL sample of ethanoic acid:
moles initial = 0.100 M ×25.0×10−3L
1= 0.00250 mol
Step 2: Identify the limiting reactant. Ethanoic acid will react with sodium
hydroxide in a 1:1 ratio. Therefore, the moles of sodium hydroxide needed to
react with the initial moles of ethanoic acid are also 0.00250 mol.
Step 3: Calculate the moles of sodium hydroxide and the remaining moles
after each volume of base is added.
a) 0.00 mL: - Moles of NaOH added: 0 - Moles of NaOH remaining: 0.00250
mol b) 12.5 mL: - Moles of NaOH added: 0.150 M ×12.5×10−3L
1= 0.00188 mol
- Moles of NaOH remaining: 0.00250 mol - 0.00188 mol = 0.00062 mol c) 25.0
mL: - Moles of NaOH added: 0.150 M ×25.0×10−3L
1= 0.00375 mol - Moles of
NaOH remaining: 0.00250 mol - 0.00375 mol = -0.00125 mol (Excess NaOH)
d) 35.0 mL: - Moles of NaOH added: 0.150 M ×35.0×10−3L
1= 0.00525 mol -
Moles of NaOH remaining: Excess NaOH e) 50.0 mL: - Moles of NaOH added:
0.150 M ×50.0×10−3L
1= 0.00750 mol - Moles of NaOH remaining: Excess NaOH
Step 4: Calculate the pH at each volume of added base. - At 0.00 mL of
NaOH added, the solution contains only ethanoic acid. Using the Henderson-
Hasselbalch equation:
pH = −log(Ka) + log [HA]=−log1.8×10−5+ log 0
0.00250= 2.74
- At 12.5 mL, 25.0 mL, 35.0 mL, and 50.0 mL of NaOH added, the solution
contains a mixture of ethanoate ions and excess sodium hydroxide. Calculate
the pH accordingly.
Question 2
Question
A 25.0 mL solution of acetic acid is titrated with 0.100 M NaOH. The pH of
the solution is monitored during the titration. At what volume of NaOH added
will the pH be equal to the pKa of acetic acid (4.74)? The Ka for acetic acid is
1.8×10−5.
Solution
Step 1: Write the chemical equation for the reaction between acetic acid and
NaOH. The reaction between acetic acid (CH3COOH) and NaOH (Na+OH−)
will produce water and sodium acetate (CH3COONa).
CH3COOH + NaOH →CH3COONa + H2O
Step 2: Calculate the initial concentration of acetic acid. The initial moles
of acetic acid is calculated using the formula:
moles of acetic acid = initial concentration ×volume
Given that volume = 25.0 mL = 0.0250 L and acetic acid is a weak acid
which does not completely dissociate, the initial concentration is 0.100 M.
2
So, moles of acetic acid = 0.100 M ×0.0250 L = 0.00250 moles
Step 3: Calculate the moles of acetic acid reacted at pH = pKa. At the
half-equivalence point, the moles of acetic acid reacted will be half of the initial
moles.
So, moles of acetic acid reacted at pH = pKa = 0.00250 moles / 2 = 0.00125
moles
Step 4: Calculate the volume of NaOH needed to reach the half-equivalence
point. The moles of NaOH required to neutralize moles of acetic acid reacted
will be the same as that of the acetic acid.
moles of NaOH = 0.00125
Using the formula:
moles of NaOH = concentration of NaOH ×volume of NaOH
We can find the volume of NaOH required.
0.00125 = 0.100 M ×volume of NaOH
volume of NaOH = 0.00125
0.100 = 0.0125 L = 12.5 mL
Therefore, the volume of NaOH added when the pH equals the pKa of acetic
acid is 12.5 mL.
Question 3
Question
A 25.0 mL solution of acetic acid (CH3COOH) of unknown concentration is
titrated with 0.100 M sodium hydroxide (NaOH). The pKa of acetic acid is 4.74.
Calculate the pH during the titration when 10.0 mL of the NaOH solution has
been added. Assume the volume of the acetic acid solution remain unchanged.
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid and sodium hydroxide.
CH3COOH + NaOH →CH3COONa + H2O
Step 2: Calculate the initial concentration of acetic acid (CH3COOH) before
any NaOH has been added.
Moles of CH3COOH = Volume ×Concentration
Moles of CH3COOH = 25.0 mL ×C
3
Where C is the initial concentration of acetic acid.
Step 3: Calculate the concentration of acetic acid after 10.0 mL of NaOH
has been added.
Moles of NaOH = Volume ×Concentration
Moles of NaOH = 10.0 mL ×0.100 M
Since the mole ratio of acetic acid to NaOH is 1:1, the moles of acetic acid
remaining is equal to the moles of NaOH reacted.
Step 4: Calculate the volume of the solution after the addition of NaOH.
25.0 mL + 10.0 mL = 35.0 mL
Step 5: Calculate the new concentration of acetic acid after the addition of
NaOH.
C=Moles of CH3COOH
Volume
C=Moles of CH3COOH
35.0 mL
Step 6: Calculate the pH after the addition of NaOH using the Henderson-
Hasselbalch equation:
pH = pKa + log [HA]
Because the solution is a buffer, we can use the Henderson-Hasselbalch equation
where [A−] is the concentration of the conjugate base (CH3COONa) and [HA]
is the concentration of the weak acid (CH3COOH).
pH = 4.74 + log CNaCH3COO
CCH3COOH
Question 4
Question
A 25.0 mL sample of hydrochloric acid solution of unknown concentration is
titrated with 0.100 M sodium hydroxide solution. The initial pH of the hy-
drochloric acid solution is 1.60. At what volume of sodium hydroxide solution
added does the equivalence point occur? The pKa of hydrochloric acid is -6.3.
Solution
Step 1: Write the balanced chemical equation for the titration. The balanced
chemical equation for the reaction between hydrochloric acid (HCl) and sodium
hydroxide (NaOH) is:
HCl(aq) + NaOH(aq)→NaCl(aq)+H2O(l)
4
Step 2: Determine the concentration of the hydrochloric acid solution. Given:
Initial pH of HCl solution = 1.60
[H+] = 10−pH = 10−1.60 = 0.025 M
Since hydrochloric acid is a strong acid that fully dissociates in solution, the con-
centration of HCl is equal to the concentration of H+ions, so the concentration
of the HCl solution is 0.025 M.
Step 3: Calculate the moles of HCl in the sample.
moles of HCl = Molarity ×Volume (L)
moles of HCl = 0.025 M ×0.0250 L = 6.25 ×10−4mol
Step 4: Determine the volume of NaOH solution required to reach the equiv-
alence point. At the equivalence point, moles of HCl = moles of NaOH. Thus,
moles of NaOH required = 6.25 x 10−4mol.
Step 5: Calculate the volume of NaOH solution required at the equivalence
point.
Volume (L) = moles
Molarity
Volume (L) = 6.25 ×10−4mol
0.100 M = 6.25 ×10−3L
Therefore, the equivalence point occurs at a volume of 6.25 mL of 0.100 M
NaOH solution.
Question 5
Question
A chemist performs a titration of a 0.020 M benzoic acid (weak acid) solution
with 0.100 M sodium hydroxide (strong base) solution. The initial volume of
benzoic acid solution is 50.0 mL. At what volume of NaOH added does the pH
of the solution equal the pKa of benzoic acid (4.20)? Assume that the volume of
NaOH added does not affect the total volume of the solution. (Kafor benzoic
acid is 6.3×10−5)
Solution
Step 1: Write the equilibrium equation for the weak acid, benzoic acid, and the
strong base, NaOH.
C6H5COOH + OH−⇌C6H5COO−+ H2O
Step 2: Calculate the initial concentrations of benzoic acid and NaOH. The
initial concentration of benzoic acid is 0.020 M. The initial concentration of
NaOH is 0.100 M.
5
Step 3: Construct an ICE table (initial, change, equilibrium) to determine
the equilibrium concentrations of the species. Let x be the amount of NaOH
(in moles) added until the pH equals the pKa. The change in concentration
of benzoic acid is -x (since it reacts with OH- in 1:1 ratio). The change in
concentration of hydroxide ion (OH-) is x. The equilibrium concentration of
benzoic acid is (0.020 - x) M. The equilibrium concentration of hydroxide ion
(OH-) is x M.
Step 4: Write the expression for the equilibrium constant, Ka.
Ka=[C6H5COO−][H2O]
[C6H5COOH]
6.3×10−5=x×x
0.020 −x
Step 5: Solve for x to find the volume of NaOH required to reach the desired
pH.
6.3×10−5=x2
0.020 −x
x2= 6.3×10−5(0.020 −x)
x2= 1.26 ×10−6−6.3×10−5x
Step 6: Rearrange the equation and solve for x using the quadratic formula.
x2+ 6.3×10−5x−1.26 ×10−6= 0
The positive root for x corresponds to the volume of NaOH added when the
pH equals the pKa.
Step 7: Calculate the pH at this volume of NaOH added. Since the pH
equals the pKa, the solution is at the halfway point of the buffer region, where
the concentrations of the weak acid and its conjugate base are equal.
pH = pKa = 4.20
Question 6
Question
A 25.0 mL sample of a weak monoprotic acid, HA, of unknown molarity is
titrated with 0.100 M NaOH. The titration curve for the reaction is shown
below.
titration_curve.png
Using the titration curve, calculate the molarity of the weak acid, HA.
6
Solution
Step 1: Identify the equivalence point from the titration curve. The equivalence
point is where the amount of acid is stoichiometrically equal to the amount of
base added. From the titration curve, we can see that the equivalence point
occurs at 25.0 mL of NaOH.
Step 2: Calculate the moles of NaOH at the equivalence point using the
formula:
moles of NaOH = M ×L
moles of NaOH = 0.100 M ×0.0250 L = 0.00250 mol
Step 3: Since the weak acid, HA, is a monoprotic acid, the moles of HA at
the equivalence point will be equal to the moles of NaOH used.
moles of HA = 0.00250 mol
Step 4: Calculate the molarity of the weak acid, HA, using the volume of
the weak acid sample and the moles of HA:
Molarity of HA = moles of HA
volume of HA
Molarity of HA = 0.00250 mol
0.0250 L = 0.100 M
Therefore, the molarity of the weak acid, HA, is 0.100 M.
Question 7
Question
Calculate the pH of a solution after 15.0 mL of 0.150 M HCl is added to 25.0
mL of 0.200 M NH3initially present. Assume the volume of the final solution
is 40.0 mL and the Kbvalue for NH3is 1.8×10−5.
Solution
Step 1: Calculate the moles of NH3and HCl initially present.
Moles of NH3= 0.0250 L ×0.200 mol/L = 0.00500 mol
Moles of HCl = 0.0150 L ×0.150 mol/L = 0.00225 mol
Step 2: Determine the limiting reactant and calculate the excess reactant.
NH3is the limiting reactant because it has fewer moles.
Excess HCl = 0.00500 mol −0.00225 mol = 0.00275 mol
7
Step 3: Calculate the moles of NH+
4and OH−formed.
Moles of NH+
4= moles of NH3used = 0.00225 mol
Moles of OH−= moles of NH3used = 0.00225 mol
Step 4: Calculate the concentration of NH+
4and OH−in the final solution.
Concentration of NH+
4=0.00225 mol
0.0400 L = 0.0563 mol/L
Concentration of OH−=0.00225 mol
0.0400 L = 0.0563 mol/L
Step 5: Calculate the concentration of NH3and HCl after the reaction.
Concentration of NH3=(0.00500 mol−0.00225 mol)
0.0400 L = 0.0563 mol/L
Concentration of HCl = 0.00275 mol
0.0400 L = 0.0688 mol/L
Step 6: Since NH3is a weak base, we need to consider the ionization of NH+
4
in water to find the pH of the solution.
Kb=[NH+
4][OH−]
[NH3]
1.8×10−5=(0.0563)(0.0563)
0.0563
[OH−]=0.0563
Step 7: Calculate the pOH and pH of the solution.
pOH =−log 0.0563 ≈1.250
pH = 14 −pOH ≈12.750
Question 8
Question
A 50.0 mL solution of 0.100 M benzoic acid (C7H6O2) is titrated with 0.200 M
NaOH. Benzoic acid is a weak monoprotic acid with a Ka value of 6.5×10−5.
Calculate the pH of the solution after the addition of 25.0 mL of the NaOH
solution.
Solution
Step 1: Write the balanced chemical equation for the reaction between benzoic
acid and NaOH. The reaction between benzoic acid (C7H6O2) and NaOH can
be written as:
C7H6O2+ NaOH →C7H5O2Na + H2O
8
Step 2: Determine the initial moles of benzoic acid. Initial moles of benzoic
acid = initial concentration ×volume (in L) Initial moles of benzoic acid =
0.100 mol/L ×50.0 mL = 0.100 mol/L ×0.050 L = 0.00500 mol
Step 3: Determine the moles of NaOH added to the solution. Moles of NaOH
added = concentration ×volume (in L) Moles of NaOH added = 0.200 mol/L
×25.0 mL = 0.200 mol/L ×0.025 L = 0.00500 mol
Step 4: Determine the limiting reactant. Since the moles of NaOH added is
equal to the initial moles of benzoic acid, NaOH is the limiting reactant.
Step 5: Calculate the moles of excess NaOH. Moles of excess NaOH = moles
of NaOH added - moles of benzoic acid Moles of excess NaOH = 0.00500 mol -
0.00500 mol = 0 mol
Step 6: Calculate the moles of benzoate ion formed. Moles of benzoate ion
formed = moles of benzoic acid reacted Moles of benzoate ion formed = 0.00500
mol
Step 7: Calculate the concentration of benzoic acid and benzoate ion. Vol-
ume after addition of NaOH = 50.0 mL + 25.0 mL = 75.0 mL = 0.075 L
Concentration of benzoate ion = moles/volume = 0.00500 mol / 0.075 L =
0.067 M Concentration of benzoic acid = 0.100 M - 0.067 M = 0.033 M
Step 8: Calculate the pH of the solution using the Henderson-Hasselbalch
equation:
pH = pKa+ log [A−]
[HA]
where [A−] is the concentration of benzoate ion and [HA] is the concentration
of benzoic acid.
pH = −log6.5×10−5+ log 0.067
0.033
pH = 4.18 + log(2) = 4.18 + 0.301 = 4.48
Therefore, the pH of the solution after the addition of 25.0 mL of the NaOH
solution is 4.48.
Question 9
Question
Calculate the pH at the equivalence point of the titration of 50.0 mL of 0.100
M acetic acid (pKa = 4.76) with 0.150 M NaOH. Assume the volume of NaOH
required to reach the equivalence point is 75.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid (CH3COOH) and sodium hydroxide (NaOH).
CH3COOH + NaOH →CH3COONa + H2O
9
Step 2: Calculate the moles of acetic acid initially present in the solution.
moles of CH3COOH = Molarity×Volume = 0.100 M×50.0×10−3L = 0.00500 mol
Step 3: Calculate the moles of NaOH added at the equivalence point.
moles of NaOH = Molarity ×Volume = 0.150 M ×75.0×10−3L=0.01125 mol
Step 4: Determine the limiting reactant. Since acetic acid reacts with NaOH
in a 1:1 ratio, moles of CH3COOH >moles of NaOH, which means NaOH is the
limiting reactant.
Step 5: Calculate the moles of excess acetic acid remaining after the reaction.
moles of CH3COOH remaining = moles ofCH3COOH initially−moles of NaOH = 0.00500 mol−0.01125 mol = −0.00625 mol
Step 6: Calculate the concentration of acetic acid at the equivalence point.
Volume of acetic acid at equivalence point = 50.0 mL+75.0 mL = 125.0 mL = 0.125 L
Concentration of CH3COOH = moles of CH3COOH remaining
Volume of CH3COOH at equivalence point =0.00625 mol
0.125 L = 0.0500 M
Step 7: Calculate the pH at the equivalence point using the Henderson-
Hasselbalch equation:
pH = pKa + log [A−]
[HA]
Plugging in the values (pKa = 4.76, [A−]=0.150 M, [HA] = 0.0500 M):
pH = 4.76 + log 0.150
0.0500= 4.76 + log(3) ≈4.76 + 0.4771 ≈5.24
Note: The negative value for moles of acetic acid remaining indicates that
all the acetic acid has been consumed at the equivalence point, leaving only the
acetate ion and water in solution.
Question 10
Question
A 25.0 mL sample of a 0.100 M weak monoprotic acid (HA) is titrated with
0.150 M NaOH. The acid dissociation constant, Ka, for HA is 1.0×10−5.
(a) Calculate the initial pH of the weak acid solution before any NaOH is
added. (b) What is the pH at the equivalence point of the titration?
10
Solution
(a) To calculate the initial pH of the weak acid solution, we need to consider
the dissociation of the weak acid and the formation of its conjugate base.
Step 1: Write the equilibrium equation for the dissociation of the weak acid:
HA ⇌H++A−
Step 2: Write the equilibrium expression for the dissociation reaction:
Ka=[H+][A−]
[HA]=x2
0.100
Step 3: Since the initial concentration of HA is 0.100 M and the initial
concentrations of H+ and A- are both 0, we can assume x mol/L of H+ and A-
are produced:
Ka=x2
0.100
Step 4: Solve for x:
x=pKa×0.100 = p(1.0×10−5)×0.100 = 3.16 ×10−3M
Step 5: Calculate the pH using the concentration of H+:
pH =−log[H+] = −log3.16 ×10−3= 2.50
Therefore, the initial pH of the weak acid solution is 2.50.
(b) At the equivalence point, the moles of base added are equal to the moles
of acid present in the solution, resulting in a solution of a weak base and its
conjugate acid.
Step 1: Determine the moles of acid initially present:
moles of acid = 0.025 L ×0.100 mol/L = 0.0025 mol
Step 2: Determine the volume of NaOH required to reach the equivalence
point:
moles of base = 0.0025 mol
volume of base = 0.0025 mol
0.150 mol/L = 0.0167 L = 16.7 mL
Step 3: Calculate the moles of NaOH left after neutralizing the acid:
moles of NaOH left = 0.025 L ×0.150 mol/L −0.0025 mol = 0.00125 mol
Step 4: Calculate the concentration of the resulting solution:
concentration of resulting solution = 0.00125 mol
0.025 L = 0.0500 mol/L
11
Step 5: Since the base (A-) is a weak base, we can consider it as a normal
weak base problem. Using the equation Kb=Kw
Ka, we can find Kb= 1.0×10−5
and Kw= 1.0×10−14:
Kb=1.0×10−14
1.0×10−5= 1.0×10−9
Step 6: Calculate the concentration of OH- ions in the solution:
OH−=sKb×concentration of resulting solution
1 + concentration of resulting solution =r(1.0×10−9)×0.0500
1+0.0500 = 5.00×10−6M
Step 7: Calculate the pOH and pH of the solution:
pOH =−log[OH−] = −log5.00 ×10−6= 5.30
pH = 14.
Question 11
Question
A 25.0 mL sample of a acetic acid (CH3COOH) solution of unknown concen-
tration is titrated with 0.150 M sodium hydroxide (NaOH). The initial pH of
the acetic acid solution is 2.85. Calculate the pH at the following volumes of
added base: (a) 0 mL, (b) 12.5 mL, (c) 25.0 mL, (d) 30.0 mL.
Given: Ka= 1.8×10−5for acetic acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid and sodium hydroxide.
CH3COOH(aq) + N aOH(aq)→CH3COON a(aq) + H2O(l)
Step 2: Calculate the initial moles of acetic acid in the solution. Initial moles
of CH3COOH =M olarity×V olume
1000 =25.0×10−3×C
1000 = 0.025C
Step 3: Calculate the initial concentration of acetic acid. Initial concentra-
tion of CH3COOH =Initial moles
V olume of solution =0.025C
25.0×10−3=C
Step 4: Calculate the initial concentration of H+ions. Since acetic acid is
a weak acid, it will only partially dissociate:
CH3COOH ⇌CH3COO−+H+
Initial concentration of H+ions = √Ka×C=√1.8×10−5×C
Step 5: Determine the initial pH of the solution.
pH =−log[H+] = −log p1.8×10−5×C=−log p1.8×10−5×C
12
Given C= 1.00 ×10−2M,
pH =−log p1.8×10−5×1.00 ×10−2=−log1.34 ×10−4= 3.87
Step 6: Calculate the moles of OH−added when 12.5 mL of N aOH is
added. Moles of N aOH =M olarity ×V olume = 0.150 ×12.5
1000 = 0.001875
Since NaOH is a strong base, all of it will dissociate to give OH−ions.
Step 7: Calculate the moles of excess OH−at 12.5 mL added. Moles of excess
OH−=Moles of NaOH −M oles of H+initially = 0.001875 −√Ka×C
Step 8: Calculate the concentration of OH−ions.
OH−=Moles of OH−
T otal volume af ter 12.5mL =0.001875 −√1.8×10−5×1.00 ×10−2
25.0 + 12.5
Step 9: Calculate the pOH at 12.5 mL added.
pOH =−log[OH−] = −log 0.001875 −√1.8×10−5×1.00 ×10−2
25.0 + 12.5!
Step 10: Calculate the pH at 12.5 mL added.
pH = 14 −pOH = 14 + log 0.001875 −√1.8×10−5×1.00 ×10−2
25.0 + 12.5!
Question 12
Question
A 25.0 mL solution of nitrous acid (HNO2) with an unknown concentration is
titrated with 0.100 M sodium hydroxide (NaOH). The initial pH of the nitrous
acid solution is 3.20. At the equivalence point, what is the pH of the solution?
The Kaof nitrous acid is 4.5×10−4.
Solution
Step 1: Write the balanced chemical equation for the reaction between nitrous
acid and sodium hydroxide:
HNO2+ NaOH →NaNO2+ H2O
Step 2: Calculate the initial concentration of nitrous acid: From the given
information, we know that the initial pH of the solution can be used to find the
initial concentration of nitrous acid using the formula:
pH = −log[H+]
13
Converting the pH to concentration:
[H+] = 10−pH = 10−3.20 = 6.31 ×10−4M
Step 3: Determine the moles of nitrous acid initially present:
mol(HNO2) = [HNO2]×volume (L) = 6.31×10−4mol/L×0.0250 L = 1.58×10−5mol
Step 4: Calculate the volume of NaOH required to reach the equivalence
point: Since the concentration of NaOH is 0.100 M and the moles of HNO2
equal the moles of NaOH at the equivalence point:
mol( NaOH) = 1.58 ×10−5mol
Volume of NaOH (L) = mol( NaOH)
concentration (M) =1.58 ×10−5mol
0.100 mol/L = 1.58×10−4L
Step 5: Calculate the concentration of NaOH at the equivalence point: The
total volume at the equivalence point is the sum of the volumes of the nitrous
acid and the NaOH:
Total volume (L) = 0.0250 L + 1.58 ×10−4L=0.0260 L
Concentration of NaOH = mol( NaOH)
Total volume (L) =1.58 ×10−5mol
0.0260 L = 6.08×10−4M
Step 6: Calculate the pH at the equivalence point using the Henderson-
Hasselbalch equation: At the equivalence point, the solution contains only the
sodium salt of the weak acid and its conjugate base. The pH is calculated using
the following formula:
pH =pKa+ log [A−]
[HA]
where [A−] is the concentration of the conjugate base (sodium nitrite) and [HA]
is the concentration of the weak acid (nitrous acid).
pKa=−log(Ka) = −log4.5×10−4= 3.35
At the equivalence point, the concentrations of the weak acid and its conju-
gate base are equal, so:
pH = 3.35 + log(1) = 3.35
Therefore, at the equivalence point, the pH of the solution is 3.35.
14
Question 13
Question
A 25.0 mL sample of 0.100 M propanoic acid (Ka = 1.3×10−5) is titrated with
0.150 M NaOH. Calculate the pH at the following points during the titration:
(a) before any NaOH is added (b) after 12.5 mL of NaOH has been added (c)
at the equivalence point (d) after 30 mL of NaOH has been added
Solution
(a) Before any NaOH is added, the solution contains only propanoic acid. To
calculate the initial pH, we need to consider the dissociation of propanoic acid.
Step 1: Write the equilibrium expression for the dissociation of propanoic
acid:
HC3H5O2 ⇌H++ C3H5O2−
Step 2: Write the expression for the acid dissociation constant, Ka:
Ka=[H+][C3H5O2−]
[HC3H5O2]
Step 3: Since propanoic acid is a weak acid, we assume that it dissociates
to a negligible extent, so we can let x be the concentration of H ions produced:
Ka=x2
0.100
Step 4: Solve for x and calculate the pH:
x=pKa×initial concentration = p(1.3×10−5)(0.100)
pH = −log(x)
(b) After 12.5 mL of NaOH has been added, we have partially neutralized
the propanoic acid. We need to calculate the number of moles of propanoic acid
and NaOH remaining.
Step 1: Calculate the number of moles of propanoic acid and NaOH initially
present.
Step 2: Calculate the number of moles of NaOH added with 12.5 mL.
Step 3: Determine the limiting reactant and calculate the remaining moles
of each species.
Step 4: Calculate the new concentrations of propanoic acid and its conju-
gate base.
Step 5: Use the Henderson-Hasselbalch equation to calculate the pH.
(c) At the equivalence point, all the propanoic acid has been neutralized by
NaOH. Calculate the pH using the concentration of the conjugate base alone.
15
(d) After 30 mL of NaOH has been added, we have excess NaOH. Calculate
the number of moles of excess NaOH and use it to determine the new concen-
trations of the species in solution. Calculate the pH using the excess hydroxide
ions.
Question 14
Question
During an acid-base titration, 25.0 mL of 0.100 M acetic acid (HC2H3O2) is
titrated with 0.150 M sodium hydroxide (NaOH). Calculate the pH at the fol-
lowing points in the titration: a) Before any NaOH is added b) Halfway to the
equivalence point c) At the equivalence point d) After the equivalence point
Given: Kafor acetic acid = 1.8×10−5
Solution
a) Before any NaOH is added: Step 1: Calculate the initial concentration of
HC2H3O2.
[HC2H3O2] = 0.100 M
Step 2: Write the equilibrium expression for the dissociation of acetic acid:
HC2H3O2⇌H++C2H3O2−
Step 3: Calculate the initial concentration of H+and C2H3O2−:
[H+] = [C2H3O2−] = 0
Step 4: Calculate the equilibrium concentration of H+using the equation
for a weak acid dissociation:
[H+]2
0.100 = 1.8×10−5
[H+]=0.00424 M
Step 5: Calculate the pH:
pH = −log(0.00424) = 2.37
b) Halfway to the equivalence point: Step 1: Calculate the moles of HC2H3O2
initially:
0.100 M ×25.0 mL = 2.50 mmol
Step 2: Calculate the moles of N aOH needed for half-neutralization:
1.25 mmol
16
Step 3: Calculate the remaining moles of HC2H3O2:
2.50 mmol −1.25 mmol = 1.25 mmol
Step 4: Calculate the concentrations of HC2H3O2 and C2H3O2−:
[HC2H3O2] = 1.25 mmol
25.0 mL = 0.050 M
Step 5: Calculate the equilibrium concentration of H+:
[H+] = p1.8×10−5×0.050 = 0.00338 M
Step 6: Calculate the pH:
pH = −log(0.00338) = 2.47
c) At the equivalence point: At the equivalence point, all the acetic acid is
neutralized by the sodium hydroxide, resulting in a solution of a sodium acetate
and water, with no acetic acid remaining.
The pH at the equivalence point will depend on the hydrolysis of the resulting
sodium acetate. The resulting solution will be basic.
d) After the equivalence point: After the equivalence point, excess sodium
hydroxide will be present in solution. The excess hydroxide ions will react with
water to form hydroxide ions and hydroxide ions in solution, resulting in a basic
solution. The pH will depend on the concentration of excess hydroxide ions.
Question 15
Question
A 35.00 mL sample of a weak monoprotic acid is titrated with 0.100 M NaOH.
The initial pH of the weak acid solution is 2.67. At what volume of NaOH added
does the pH of the solution reach 8.00? Assume that the volume of NaOH added
is less than the equivalence point volume.
Solution
Step 1: Write the balanced chemical equation for the reaction between the weak
monoprotic acid (HA) and NaOH:
HA +NaOH →N aA +H2O
Step 2: Calculate the initial concentration of the weak acid (HA) using the
given initial pH:
[HA] = 10−pH = 10−2.67 = 2.06 ×10−3M
17
Step 3: Set up an ICE table (Initial, Change, Equilibrium) to track the
concentrations of the weak acid (HA), its conjugate base (A), and the added
OH ions as NaOH is titrated.
Step 4: At the halfway point to the equivalence point, the pH of the solution
will be equal to the pKa of the weak acid. Use the Henderson-Hasselbalch
equation to find the concentration of A at this point:
pH = pKa + log [A−]
[HA]
8.00 = pKa + log [A−]
2.06 ×10−3
Step 5: From the ICE table, the initial concentration of A is 0, and the initial
concentration of HA is 2.06 ×10−3M. Let x be the change in concentration of
HA and A as the titration progresses to the halfway point:
HA →A−+ H+
2.06 ×10−3−xxx
Step 6: Substitute the concentrations into the Henderson-Hasselbalch equa-
tion and solve for x:
8.00 = pKa + log x
2.06 ×10−3−x
Step 7: Once x is found, calculate the volume of NaOH added up to that
point using the stoichiometry of the reaction and the molarity of NaOH.
Question 16
Question
A 25.0 mL sample of a weak monoprotic acid with an unknown concentration
is titrated with 0.150 M NaOH. The initial pH of the weak acid solution is 3.85.
At what volume of NaOH added does the pH change the most rapidly? Justify
your answer.
Solution
Step 1: Write the balanced chemical equation for the reaction between the weak
acid and NaOH. The reaction between a weak monoprotic acid (HA) and NaOH
can be represented as:
HA(aq) + N aOH(aq)→N aA(aq) + H2O(l)
Step 2: Determine the equivalence point of the titration. The equivalence
point occurs when moles of acid = moles of base. Since we have a weak mono-
protic acid, the moles of acid can be determined using the initial concentration
18
and volume of the acid, and the moles of base can be determined using the
volume and concentration of NaOH. By equating these two quantities, we can
find the volume of NaOH required to reach the equivalence point.
Step 3: Plot the titration curve. Sketch a rough graph of the titration curve
showing how the pH changes as NaOH is added. The pH initially rises slowly as
NaOH is added due to the buffering capacity of the weak acid, then rises more
rapidly near the equivalence point, and finally levels off after the equivalence
point is reached.
Step 4: Analyze the pH changes. The pH changes most rapidly around the
equivalence point because that is where the concentration of the weak acid is
changing most rapidly. At the equivalence point, the rate of change of pH is
maximum.
Therefore, the volume of NaOH added that causes the pH to change most
rapidly is at the equivalence point.
Question 17
Question
A solution contains acetic acid (CH3COOH) with a concentration of 0.100 M.
It is titrated with a 0.150 M sodium hydroxide (NaOH) solution. Calculate the
pH of the solution at the following points of the titration: a) Before any NaOH
is added. b) After 20.0 mL of N aOH has been added. c) At the equivalence
point. d) After 30.0 mL of N aOH has been added after the equivalence point.
Given the Kaof acetic acid is 1.8×10−5.
Solution
Step 1: Before any NaOH is added.
Since no NaOH has been added yet, the solution only contains acetic acid.
We can assume that the solution will be a buffer solution with acetic acid and its
conjugate base, acetate (CH3COO−). The initial concentration of CH3COOH
is 0.100 M.
The initial moles of acetic acid is:
0.100 M ×Vtotal = 0.100 M ×50.0×10−3L=5.00 ×10−3mol
So, the initial moles of CH3COO−(acetate) will also be 5.00 ×10−3mol
(since acetic acid fully dissociates to produce acetate and H+).
From this information, we can calculate the initial pH of the solution before
any NaOH is added. Since we’re dealing with a buffer solution, Henderson-
Hasselbalch equation can be used:
pH =pKa + log [A−]
[HA]
19
where [A−] is the concentration of acetate and [HA] is the concentration of
acetic acid.
pH =−log1.8×10−5+ log 5.00 ×10−3
5.00 ×10−3
=−(−4.74) + log(1)
= 4.74 + 0
= 4.74
Thus, the pH before any NaOH is added is 4.74.
Question 18
Question
A 25.0 mL solution of acetic acid (CH3COOH) with an unknown concentration
is titrated with 0.100 M sodium hydroxide (NaOH). The pKa of acetic acid is
4.76. At what volume of NaOH added does the pH of the solution equal the
pKa of acetic acid?
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid and sodium hydroxide.
CH3COOH + OH−→CH3COO−+ H2O
Step 2: Calculate the initial moles of acetic acid in the solution.
moles of CH3COOH = M ×L=C1×V1
moles of CH3COOH = C1×V1= C1×V1
moles of CH3COOH = (C1)(V1)=(xM)(0.0250 L) = 0.025xmoles
Step 3: Since acetic acid is a weak acid, it undergoes partial dissociation
according to the reaction equation. Let the change in moles of acetic acid be -x
moles. This is because x moles of acetic acid reacts with x moles of NaOH.
The moles of acetic acid left after the reaction is (0.025x - x) = 0.025x - x
= 0.025 - 1.
The moles of acetate ion formed (0.025x + x) = 0.ml025x + x = 0.025 + 1
= 0.025 + 1.
Step 4: Calculate the concentration of acetic acid and acetate ion.
CCH3COOH = moles of CH3COOH remaining
total volume
CCH3COO−=moles of CH3COO−f ormed
total volume
20
CCH3COOH = 0.025 −x
0.025 + 0.025 =0.025 −1
0.05
CCH3COO−=0.025x+x
0.05 =0.025 + 1
0.05
Step 5: Calculate the concentrations of acetic acid, acetate ion, and hydrox-
ide ion at the equivalence point.
At the equivalence point, all the acetic acid has been neutralized by the
sodium hydroxide. Therefore, the moles of acetic acid equals the moles of sodium
hydroxide added.
CCH3COOHVCH3COOH = CNaOHVNaOH
(0.025)(0.100) ×V= 0.100 ×V
V=0.025 moles
0.100 M = 0.025 L = 25.0 mL
Therefore, the volume of NaOH added when the pH equals the pKa of acetic
acid is 25.0 mL.
Question 19
Question
A weak monoprotic acid, HA, with a dissociation constant of 1.5×10−5, is
titrated with a 0.10 M solution of NaOH. Calculate the pH at the following
volumes of NaOH added:
0 mL
25 mL
50 mL
75 mL
100 mL
Assume a total volume of 100 mL and ignore volume changes due to mixing.
Solution
Step 1: Calculate the initial moles of acid. The initial moles of acid can be
calculated as:
moles of acid = initial concentration ×volume
21
At 0 mL, the initial moles of acid are:
moles of acid at 0 mL = 0.10 M ×100 mL = 10 mmol
Step 2: Determine the moles of acid and base reacted at each volume of
NaOH added. At 25 mL, 50 mL, 75 mL, and 100 mL, the moles of NaOH added
can be calculated as:
moles of NaOH = 0.10 M ×volume added
Using stoichiometry, we can determine the moles of acid and base reacted.
Step 3: Calculate the final moles of acid at each volume of NaOH added.
The final moles of acid can be determined by subtracting the moles of base
reacted from the initial moles of acid.
Step 4: Calculate the equilibrium concentration of HA at each volume of
NaOH added. The equilibrium concentration of the weak acid can be expressed
in terms of the initial moles of acid and the moles of acid reacted, as well as the
total volume.
Step 5: Calculate the pH at each volume of NaOH added using the Henderson-
Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log [A−]
[HA]
where [A−] is the concentration of the conjugate base and [HA] is the concen-
tration of the weak acid (HA).
Question 20
Question
A 25.00 mL sample of 0.100 M acetic acid (CH3COOH) is titrated with 0.100 M
sodium hydroxide (NaOH). Calculate the pH at the following volumes of added
base: (a) 0 mL (b) 12.50 mL (c) 25.00 mL (d) 50.00 mL
Given: Ka= 1.8×10−5for acetic acid
Solution
Step 1: Calculate the initial moles of acetic acid.
moles of CH3COOH = concentration×volume = 0.100 M×25.00×10−3L=0.00250 mol
Step 2: Calculate the moles of sodium hydroxide. (a) For 0 mL of added
base: In this case, moles of CH3COOH = moles of OH−.
0.00250 mol −0=0.00250 mol
22
Step 3: Calculate the concentrations of the acids and bases. (a) For 0 mL
of added base:
[CH3COOH] = 0.00250 mol
25.00 ×10−3L= 0.100 M
[OH−] = 0 mol
25.00 ×10−3L= 0 M
[H+] = [OH−] = 0 M
Since Kw= [H+][OH−] = 1.0×10−14, we have:
0=0.100 ×10−14 = 1.0×10−14
Therefore, the pH is 7 for 0 mL of added base.
(b) For 12.50 mL of added base: The procedure for calculating the pH at
12.50 mL, 25.00 mL, and 50.00 mL would involve a similar series of steps as
above. Would you like to continue with the calculations for parts (b), (c), and
(d)?
Question 21
Question
A 25.0 mL solution of acetic acid with a concentration of 0.100 M is titrated
with 0.150 M NaOH. At what volume of NaOH added (in mL) will the pH of
the solution be equal to the pKa of acetic acid (4.76)? The Ka value for acetic
acid is 1.8×10−5.
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid (CH3COOH) and NaOH (Na+OH−).
CH3COOH + NaOH →CH3COONa + H2O
Step 2: Calculate the initial moles of acetic acid and NaOH.
moles of CH3COOH = concentration ×volume
= 0.100 M ×25.0×10−3L
= 2.50 ×10−3mol
moles of NaOH = concentration ×volume
= 0.150 M ×VL
= 0.150Vmol
23
Step 3: Determine the limiting reactant and the moles of the excess reactant.
Since CH3COOH and NaOH react in a 1:1 ratio, the limiting reactant is the
one that is completely used up first, which is the one with the smaller number
of moles. In this case, the limiting reactant is CH3COOH, as all of it is used
up first. The moles of NaOH remaining after the reaction with CH3COOH is
0.150V−2.50 ×10−3mol.
Step 4: Calculate the concentration of acetic acid, acetate ion, and [H+]
after the reaction. The final volume after NaOH reaction is 25.0 mL + V mL.
The concentration of acetate ion:
[CH3COO−] = 2.50 ×10−3mol
25.0×10−3+VL
The concentration of acetic acid:
[CH3COOH] = 2.50 ×10−3mol
25.0×10−3+VL
The concentration of [H+]:
[H+] = Ka[CH3COO−]
[CH3COOH] =1.8×10−5×2.50 ×10−3
25.0×10−3+V
Step 5: Considering the equation for the dissociation of acetic acid, deter-
mine when [H+] equals 10−4.76 M.
1.8×10−5×2.50 ×10−3
25.0×10−3+V= 10−4.76
Step 6: Solve the equation for V to find the volume of NaOH added when
the pH equals the pKa of acetic acid.
Question 22
Question
A 25.0 mL sample of a weak monoprotic acid with an initial concentration of
0.100 M is titrated with 0.150 M NaOH. At the equivalence point, the pH of
the solution is 10.84. Calculate the pKa of the acid.
Solution
Step 1: Write the reaction that occurs during the titration. The reaction be-
tween the weak monoprotic acid (HA) and NaOH can be written as:
HA +NaOH →N aA +H2O
Step 2: Determine the initial moles of the acid.
Initial moles of HA = initial concentration ×volume
24
= 0.100 M ×25.0 mL ×1 L
1000 mL
= 0.00250 mol
Step 3: Use the stoichiometry of the reaction to determine the moles of
NaOH added at the equivalence point. Since the acid is monoprotic, the moles
of NaOH added will be equal to the moles of HA initially present. Therefore,
at the equivalence point, the moles of NaOH added = 0.00250 mol.
Step 4: Calculate the volume of NaOH added at the equivalence point.
Volume of NaOH added = moles
molarity
=0.00250 mol
0.150 M
= 16.7 mL
Step 5: Calculate the moles of excess NaOH at the equivalence point.
Excess moles of NaOH = total moles of NaOH added −moles of HA
= 0.00250 mol −0.00250 mol
= 0 mol
Step 6: Calculate the concentration of excess OH−ions at the equivalence
point.
Volume of excess NaOH = 25.0 mL −16.7 mL
= 8.3 mL = 0.0083 L
Concentration of excess OH−=moles of excess NaOH
volume of excess NaOH
=0 mol
0.0083 L
= 0 M
Step 7: Calculate the pKa value of the acid using the formula:
pKa = pH at equivalence point + log Concentration of excess OH−
Initial concentration of HA
pKa = 10.84 + log 0
0.100
pKa = 10.84 + log(0)
Since the logarithm of zero is undefined, it means that the acid is completely
titrated at the equivalence point and there is no excess acid left to determine
the pKa.
25
Question 23
Question
A 25.0 mL solution of acetic acid is titrated with 0.100 M sodium hydroxide.
The pKa of acetic acid is 4.74. Calculate the pH of the solution at the following
points during the titration: a) Before any NaOH is added b) After the addition
of 12.5 mL of NaOH c) At the equivalence point d) After the addition of 35.0
mL of NaOH
Solution
a) Before any NaOH is added: The initial pH of the acetic acid solution can be
calculated using the formula for weak acids:
pH = 1
2(pKa −log [acid])
Substitute the given values:
pH = 1
2(4.74 −log [acid])
b) After the addition of 12.5 mL of NaOH: The number of moles of acetic
acid initially present can be calculated:
mol of acetic acid = Molarity ×Volume (L)
Then, calculate the number of moles of NaOH added:
mol of NaOH = Molarity ×Volume (L)
Determine which reactant is limiting and calculate the excess amount and re-
sulting pH.
c) At the equivalence point: At the equivalence point, all the acetic acid
has reacted with the NaOH to form sodium acetate and water. Calculate the
resulting pH using the pKb of the conjugate base.
d) After the addition of 35.0 mL of NaOH: Similarly to part (b), calculate
the excess amount of NaOH and the resulting pH after reaction with the acetic
acid.
Question 24
Question
A 25.00 mL sample of a weak monoprotic acid of unknown concentration is
titrated with 0.150 M NaOH. The initial pH of the acid is 2.80. The equivalence
point is reached after adding 22.50 mL of the NaOH solution. Calculate the pKa
of the acid.
26
Solution
Step 1: Calculate the initial moles of the acid. Given: Volume of acid sample =
25.00 mL, initial pH = 2.80. The initial concentration of H+ions can be calcu-
lated using the formula pH = -log[H+]. Thus, [H+] = 10−pH mol/L. Substitute
the given pH value into the formula: [H+] = 10−2.80 mol/L. The initial moles
of the acid can be calculated using: moles of the acid = initial concentration ×
volume. moles of the acid = [H+]×(volume of acid sample in L). moles of the
acid = 10−2.80 mol/L ×0.02500 L.
Step 2: Calculate the moles of NaOH at the equivalence point. Given:
Volume of NaOH added = 22.50 mL, concentration of NaOH = 0.150 M. The
moles of NaOH used at the equivalence point can be calculated using: moles of
NaOH = concentration ×volume. moles of NaOH = 0.150 mol/L ×0.02250 L.
Step 3: Determine the moles of the acid left at the equivalence point. Since
the moles of acid and moles of base are equal at the equivalence point, the moles
of acid remaining can be calculated. moles of acid remaining = initial moles of
acid - moles of NaOH. moles of acid remaining = 10−2.80 mol/L ×0.02500 L -
0.150 mol/L ×0.02250 L.
Step 4: Calculate the volume of NaOH needed to reach the half-equivalence
point. At the half-equivalence point, half of the moles of acid have reacted.
Thus, moles of acid at half-equivalence point = initial moles of acid / 2. The
volume of NaOH needed to reach the half-equivalence point can be calculated
from the moles of NaOH at the equivalence point and the moles of acid at
half-equivalence point.
Step 5: Determine the pKa of the acid. The pKa of the acid can be calculated
using the Henderson-Hasselbalch equation: pKa = pH + log[A−]
[HA] . In this
case, at the half-equivalence point, [A−] = [HA]. Substitute the pH at the half-
equivalence point and the concentration of the acid to calculate the pKa.
Question 25
Question
A 0.100 M solution of a monoprotic weak acid, HA, with a pKa of 4.75, is
titrated with 0.200 M NaOH. Calculate the pH of the solution after 50.0 mL of
NaOH has been added to 25.0 mL of the weak acid solution.
Solution
Step 1: Calculate the initial moles of HA present in the solution. Given: Volume
of HA solution = 25.0 mL = 0.025 L Concentration of HA = 0.100 M
Initial moles of HA = concentration ×volume = 0.100 mol/L ×0.025 L =
0.0025 moles
Step 2: Calculate the moles of NaOH added. Given: Volume of NaOH
solution added = 50.0 mL = 0.050 L Concentration of NaOH = 0.200 M
27
Moles of NaOH = concentration ×volume = 0.200 mol/L ×0.050 L = 0.010
moles
Step 3: Determine the limiting reactant in the reaction between HA and
NaOH. Since HA is a weak acid and NaOH is a strong base, a neutralization
reaction occurs between them. The limiting reactant will be the one that is
completely consumed, which in this case is HA.
Step 4: Calculate the moles of excess NaOH after the reaction is complete.
Moles of excess NaOH = Moles of NaOH added - Moles of HA reacted = 0.010
moles - 0.0025 moles = 0.0075 moles
Step 5: Calculate the final volume of the solution. Final volume = volume
of HA solution + volume of NaOH added = 0.025 L + 0.050 L = 0.075 L
Step 6: Calculate the concentration of the excess NaOH in the final solution.
Concentration of excess NaOH = moles of excess NaOH / final volume = 0.0075
moles / 0.075 L = 0.100 M
Step 7: Calculate the moles of OH- ions in the final solution. Since NaOH
completely dissociates in solution, moles of OH- = moles of excess NaOH =
0.0075 moles
Step 8: Calculate the concentration of OH- ions in the final solution. Con-
centration of OH- = moles of OH- / final volume = 0.0075 moles / 0.075 L =
0.100 M
Step 9: Calculate the pOH of the final solution. pOH = - log(OH-) = -
log(0.100) = 1.00
Step 10: Calculate the pH of the final solution. pH + pOH = 14 pH = 14 -
pOH = 14 - 1.00 = 13.00
Therefore, the pH of the solution after 50.0 mL of NaOH has been added to
25.0 mL of the weak acid solution is 13.00.
Question 26
Question
A 50.0 mL solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH.
The pH at different points during the titration is given in the table below. Use
this data to draw a titration curve and determine the Kafor acetic acid.
Volume of NaOH added (mL) pH
0 2.87
10 4.41
20 4.76
30 4.94
40 5.06
50 5.14
28
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid and sodium hydroxide.
CH3COOH + NaOH →CH3COONa + H2O
Step 2: Calculate the initial concentration of acetic acid. Since we have a
50.0 mL solution of acetic acid, the initial moles of CH3COOH is:
nCH3COOH =50.0 mL
1000 mL/L ×0.100 mol/L = 0.00500 mol
Therefore, the initial concentration of CH3COOH is:
cCH3COOH =0.00500 mol
0.0500 L = 0.100 mol/L
Step 3: Construct the titration curve using the given pH values. The titra-
tion curve will show the pH plotted against the volume of NaOH added. We
will plot the data points given in the table and connect them to form a curve.
Step 4: Determine the equivalence point. The equivalence point is where
the moles of CH3COOH are equal to the moles of NaOH added. By using the
initial moles of acetic acid and the volume of NaOH added at the equivalence
point, determine the Kavalue for acetic acid.
Question 27
Question
A 25.00 mL sample of a 0.150 M solution of an unknown monoprotic weak acid
is titrated with 0.100 M NaOH. The initial pH of the solution is 2.90. Calculate
the pH when 20.00 mL of NaOH has been added to the solution. The pKa of
the weak acid is 3.75.
Solution
Step 1: Calculate the initial concentration of the weak acid. Given: Volume of
weak acid solution (Va) = 25.00 mL = 0.025 L Molarity of weak acid solution
(Ma) = 0.150 M
Initial moles of weak acid = Ma×Va= 0.150 M ×0.025 L = 0.00375 mol
Step 2: Determine the moles of NaOH added. Volume of NaOH added (Vb)
= 20.00 mL = 0.020 L Molarity of NaOH (Mb) = 0.100 M
Moles of NaOH = Mb×Vb= 0.100 M ×0.020 L = 0.002 mol
29
Step 3: Identify the limiting reactant and calculate the excess. Since NaOH
is a strong base, it completely reacts with the weak acid.
Moles of weak acid remaining = 0.00375 mol −0.002 mol = 0.00175 mol
Step 4: Calculate the volume of the weak acid solution after the addition of
NaOH.
Vfinal =Va−Vb= 0.025 L −0.020 L = 0.005 L
Step 5: Calculate the concentration of the weak acid after the addition of
NaOH.
Mfinal =moles of weak acid remaining
Vfinal
=0.00175 mol
0.005 L = 0.350 M
Step 6: Calculate the H+concentration to find the pH. The weak acid
partially dissociates. Let xbe the amount of H+ions dissociated.
Ka =x2
Mfinal
= 10−pKa = 10−3.75
Solving for xgives x= 0.0245 M. Thus, [H+] = 0.0245M.
Step 7: Calculate the pH of the solution.
pH =−log[H+] = −log(0.0245) = 1.61
Therefore, the pH of the solution after 20.00 mL of NaOH has been added
is 1.61.
Question 28
Question
A 25.0 mL solution containing 0.100 M benzoic acid (C6H5COOH) is titrated
with 0.200 M NaOH. The pKaof benzoic acid is 4.19. Calculate the pH at the
following volumes of added NaOH: (a) 0 mL, (b) 5.00 mL, (c) 10.0 mL, (d) 12.5
mL, and (e) 25.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between benzoic
acid and NaOH.
C6H5COOH (aq) + NaOH (aq) →C6H5COONa (aq) + H2O (l)
Step 2: Calculate the initial concentration of benzoic acid and NaOH in the
solution. Initial moles of benzoic acid = (25.0 mL)(0.100 mol/L) = 2.50 mmol
Initial moles of NaOH = 0 mmol
30
Step 3: Determine the limiting reactant and moles of benzoate ion formed.
Given that benzoic acid reacts with NaOH in a 1:1 ratio, and NaOH is the
limiting reactant, the moles of benzoic acid will be fully converted to benzoate
ion. Moles of benzoate ion formed = 2.50 mmol
Step 4: Calculate the total volume of the solution after each volume of NaOH
is added. (a) For 0 mL NaOH added: 25.0 mL + 0 mL = 25.0 mL (b) For 5.00
mL NaOH added: 25.0 mL + 5.00 mL = 30.0 mL (c) For 10.0 mL NaOH added:
25.0 mL + 10.0 mL = 35.0 mL (d) For 12.5 mL NaOH added: 25.0 mL + 12.5
mL = 37.5 mL (e) For 25.0 mL NaOH added: 25.0 mL + 25.0 mL = 50.0 mL
Step 5: Determine the final concentration of benzoic acid and benzoate ion in
the solution after each volume of NaOH is added. (a) For 0 mL NaOH added: -
Concentration of benzoic acid = 2.50 mol
25.0 mL = 0.100 M - Concentration of benzoate
ion = 0 M
(b) For 5.00 mL NaOH added: - Concentration of benzoic acid = 2.50 mmol
30.0 mL =
0.083 M - Concentration of benzoate ion = 2.50 mmol
30.0 mL = 0.083 M
Question 29
Question
The titration of 50.00 mL of 0.10 M acetic acid (CH3COOH) with 0.10 M
sodium hydroxide (NaOH) has an initial pH of 2.85. Calculate the pH of the
solution after the addition of 25.00 mL of the sodium hydroxide solution. (Ka
of acetic acid = 1.8×10−5)
Solution
Step 1: Calculate the initial moles of acetic acid. Given: Initial volume of acetic
acid = 50.00 mL Initial concentration of acetic acid = 0.10 M
Using the formula:
moles = volume ×concentration
we have:
moles of acetic acid = 50.00 ×0.10 = 5.00 mmol
Step 2: Determine the moles of N aOH added. The volume of NaOH added
is 25.00 mL or 0.025 L.
moles of NaOH = volume ×concentration
moles of NaOH = 0.025 ×0.10 = 0.0025 mol
Step 3: Calculate the moles of acetic acid and N aOH remaining after the
reaction. Since the reaction between CH3COOH and N aOH is stoichiometric
with a 1:1 ratio, the limiting reagent will be the one that is completely consumed.
In this case, acetic acid will be completely consumed.
31
Moelar ratio of CH3COOH :N aOH = 1:1
Therefore, moles of CH3COOH remaining after the reaction = 5.00 - 0.0025
= 4.9975 mmol.
Step 4: Calculate the concentration of acetic acid and N aOH in the final
volume. The final volume after adding 25.00 mL of N aOH will be 75.00 mL or
0.075 L.
Concentration of acetic acid:
Final concentration of CH3COOH =4.9975 mmol
0.075 L = 0.0666 M
Concentration of NaOH: Since 0.0025 mol of N aOH was added to 0.050 L
of the acetic acid solution, the concentration of the NaOH in the final solution
is:
Final concentration of NaOH =0.0025 mmol
0.075 L = 0.03333 M
Step 5: Calculate the pH of the buffer solution. The pH of a weak acid buffer
solution can be calculated using the Henderson-Hasselbalch equation:
pH =pKa + log [A−]
[HA]
where: [A−] is the concentration of the conjugate base (CH3COO−) [HA] is
the concentration of the weak acid (CH3COOH)
The equilibrium concentrations at the half-equivalence point will be equal
for CH3COOH and CH3COO−, so [CH3COO−] = [CH3COOH].
Substitute the values into the Henderson-Hasselbalch equation:
pH =−log1.8×10−5+ log(1) = −(log(1.8) + log10−5) = 4.74
Therefore, the pH of the solution after adding 25.00 mL of the sodium hy-
droxide solution is 4.74.
Question 30
Question
A 25.0 mL sample of hydrochloric acid solution with an unknown concentration
was titrated with 0.100 M sodium hydroxide solution. The titration curve shown
below was obtained.
titration_curve.png
Determine the concentration of the hydrochloric acid solution.
32
Solution
To determine the concentration of the hydrochloric acid solution, we need to
find the equivalence point and use the volume of sodium hydroxide solution
added at that point.
Step 1: Determine the equivalence point From the titration curve,
we can see that the equivalence point occurs where the pH undergoes a sharp
increase. In this case, the equivalence point is around 25.0 mL.
Step 2: Calculate the moles of sodium hydroxide at the equivalence
point At the equivalence point, moles of acid = moles of base
moles of NaOH = MNaOH ×VNaOH
moles of NaOH = 0.100 M ×25.0×10−3L
moles of NaOH = 0.00250 mol
Step 3: Calculate the moles of hydrochloric acid at the equivalence
point Since hydrochloric acid reacts with sodium hydroxide in a 1:1 ratio:
moles of HCl = moles of NaOH
moles of HCl = 0.00250 mol
Step 4: Calculate the concentration of hydrochloric acid Now, we
can determine the concentration of the hydrochloric acid solution using the
volume of the acid titrated and the moles of the acid at the equivalence point:
MHCl =moles of HCl
VHCl
MHCl =0.00250 mol
25.0×10−3L
MHCl = 0.10 M
Therefore, the concentration of the hydrochloric acid solution is 0.10 M.
Question 31
Question
A 50.0 mL sample of a monoprotic weak acid, HA, with an unknown concentra-
tion is titrated with 0.100 M NaOH. The initial pH of the weak acid solution is
2.60. At the equivalence point, the pH is 10.50. Calculate the pKa of the weak
acid HA.
33
Solution
Step 1: Write the balanced chemical equation for the reaction between the weak
acid HA and NaOH.
HA(aq) + OH−(aq)→A−(aq)+H2O(l)
Step 2: Determine the initial concentration of the weak acid HA. From the
initial pH of 2.60, we can calculate the initial concentration of HA using the
formula:
pH = pKa + log [A−]
[HA]
Since the weak acid is monoprotic, initially, [HA] = [HA]0. Therefore, we have:
2.60 = pKa + log 0
[HA]0= pKa −log([HA]0)
⇒[HA]0= 10−pKa+2.60
Step 3: Calculate the volume of NaOH needed to reach the equivalence point.
Since the weak acid is monoprotic, the moles of OH−added at the equivalence
point will be equal to the initial moles of HA.
moles of HA = volume of NaOH at equivalence point ×concentration of NaOH
[HA]0×Vequiv = 0.100 M ×Vequiv
[HA]0= 0.100 M
Step 4: Write the equation for the weak acid dissociation reaction.
HA(aq)⇌H+(aq)+A−(aq)
Step 5: Write the equilibrium expression for the dissociation of the weak
acid HA.
Ka=[H+][A−]
[HA]0
Step 6: Use the equilibrium expression to find the pKa of the weak acid HA.
Since [A−] = [HA]0at the equivalence point,
Ka= [H+]×[HA]0
At the equivalence point, [H+] = 10−pOH = 10−10.50. Therefore,
10−10.50 ×0.100 = 10−pKa+2.60
Now solve for pKa to determine the pKa of the weak acid HA from the titration
curve data.
34
Question 32
Question
A 25.0 mL sample of 0.100 M acetic acid (CH3COOH) is titrated with 0.100
M sodium hydroxide (NaOH). Calculate the pH of the solution at the following
points of the titration:
1. Before any NaOH is added.
2. At the equivalence point.
3. After 20.0 mL of NaOH is added.
Given: Kaof acetic acid = 1.8×10−5
Solution
Given that the initial concentration of acetic acid is 0.100 M, the initial concen-
tration of CH3COO−is 0 M and the initial concentration of H+ions is also 0
M. Let’s calculate the pH at each point of the titration.
Before any NaOH is added:
[CH3COOH] = 0.100 M
[CH3COO−] = 0 M
[H+] = 0 M
The pH of acetic acid at the start is the pH of a weak acid:
pH = 1
2(pKa−log[CH3COOH]) = 1
2(4.74 −log0.100) = 2.87
At the equivalence point: At the equivalence point, the moles of acetic
acid equal the moles of NaOH:
moles of acetic acid = 0.025 L ×0.100 M
= 0.0025 mol
Since acetic acid is a monoprotic acid, 0.0025 mol of NaOH is required to
reach the equivalence point. This results in a solution of 0.050 L of 0.050 M
CH3COONa. We can use the Henderson-Hasselbalch equation to calculate the
pH at the equivalence point:
pH = pKa+ log [CH3COO−]
[CH3COOH]= 4.74 + log 0.050
0.050= 4.74
After 20.0 mL of NaOH is added: After adding 20.0 mL of 0.100 M
NaOH, we have 0.0020 mol of NaOH in solution. This reacts with acetic acid
to form CH3COONa:
moles of acetic acid = 0.025 L ×0.100 M −0.020 L ×0.100 M
= 0.0005 mol
35
This results in a solution of CH3COOH and CH3COO−, which can be used with
the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa+ log [CH3COO−]
[CH3COOH]= 4.74 + log 0.020
0.025= 4.45
Question 33
Question
A 25.0 mL sample of a weak monoprotic acid, HX, is titrated with 0.100 M
NaOH. The initial pH of the acid solution is 3.20. At the equivalence point, the
pH is 11.90. Calculate the pKa of the acid.
Solution
Step 1: Write the balanced chemical equation for the reaction between the weak
acid and NaOH. The reaction between a weak monoprotic acid and a strong base
can be represented as:
HX(aq) + N aOH(aq)→N aX(aq) + H2O(l)
Step 2: Determine the initial concentration of the weak acid, HX. From the
given initial pH of 3.20, we can calculate the initial concentration of HX using
the formula:
pH =−log[H3O+]
3.20 = −log[H3O+]
[H3O+] = 10−3.20
Step 3: Determine the moles of HX in the solution.
[HX] = moles
volume(L)
[HX] = moles
0.0250L
Step 4: Determine the moles of NaOH added at the equivalence point. At
the equivalence point, moles of NaOH = moles of HX. Therefore, moles of NaOH
added = moles of HX initially present.
Step 5: Calculate the concentration of HX at the equivalence point. At the
equivalence point, the volume of the solution is doubled to 50.0 mL. Use the
moles of HX (from Step 4) and the total volume to calculate the concentration
of HX.
Step 6: Calculate the pKa of the acid. The Henderson-Hasselbalch equation
is given by:
pH =pKa + log [A−]
[HA]
36
At the equivalence point, the pH is equal to 11.90. Substitute the concentration
of HX and the concentration of NaX into the Henderson-Hasselbalch equation
to solve for pKa.
Question 34
Question
A 25.0 mL solution of acetic acid, CH3COOH, of unknown concentration is
titrated with 0.100 M NaOH. The pKa of acetic acid is 4.76. At what volume
of NaOH added would the pH be equal to the pKa of acetic acid?
Solution
Step 1: Write the balanced chemical equation for the reaction between acetic
acid and NaOH. The balanced chemical equation for the reaction is: CH3COOH
+ NaOH →CH3COONa + H2O
Step 2: Calculate the initial moles of acetic acid in the solution. Given that
the volume of acetic acid solution is 25.0 mL and the concentration is unknown,
let’s denote it as C (in mol/L). The initial moles of acetic acid = C ×25.0 mL
=C
1000 ×25.0 mol
Step 3: Determine the moles of NaOH added when the pH = pKa. At the
equivalence point, the moles of acetic acid = moles of NaOH added. Initially,
the moles of acetic acid = C
1000 ×25.0 mol Since the concentration of NaOH is
0.100 M, moles of NaOH added = 0.100 mol/L ×volume added (in L). At the
equivalence point, moles of acetic acid = moles of NaOH added: C
1000 ×25.0 =
0.100 ×volume added
Step 4: Calculate the volume of NaOH added at the equivalence point when
the pH = pKa. Given that at the equivalence point the pH = pKa, we can use the
Henderson-Hasselbalch equation: pH =pKa + log [A−]
[HA]At the equivalence
point, [A−] = concentration of the acetate ion = moles of NaOH added (in mol)
/ total volume (in L) [HA] = concentration of acetic acid = (moles of acetic
acid remaining + moles of acetic acid initially) / total volume (in L) Since
moles of acetic acid initially = moles of NaOH added at the equivalence point:
Volume of NaOH added = C
1000 ×25.0 mL
Therefore, at what volume of NaOH added would the pH be equal to the
pKa of acetic acid? The volume of NaOH added would be C
1000 ×25.0 mL
Question 35
Question
A 50.00 mL solution of phosphoric acid (H3PO4) of unknown concentration is
titrated with 0.1000 M NaOH. The plot of pH versus volume of NaOH added is
shown below.
37
titration_curve.png
Determine the concentration of the phosphoric acid solution.
Solution
Step 1: Identify the equivalence points on the titration curve. The titration
curve shows three equivalence points corresponding to the three acidic protons
in phosphoric acid (H3PO4). The first equivalence point occurs at around 10
mL (corresponding to the H3P O4→H2P O−
4transition), the second equiva-
lence point at around 25 mL (corresponding to the H2P O−
4→HP O2−
4transi-
tion), and the third equivalence point at around 40 mL (corresponding to the
HP O2−
4→P O3−
4transition).
Step 2: Calculate the number of moles of NaOH at each equivalence point.
At the first equivalence point: - Moles of NaOH = volume of NaOH (L) ×
concentration of NaOH (mol/L) - Moles of NaOH = 0.010 L ×0.100 mol/L =
0.001 mol
At the second equivalence point: - Moles of NaOH = volume of NaOH (L)
×concentration of NaOH (mol/L) - Moles of NaOH = 0.025 L ×0.100 mol/L
= 0.0025 mol
At the third equivalence point: - Moles of NaOH = volume of NaOH (L) ×
concentration of NaOH (mol/L) - Moles of NaOH = 0.040 L ×0.100 mol/L =
0.004 mol
Step 3: Calculate the moles of phosphoric acid at each equivalence point
using the mole ratios from the balanced chemical equation. Since phosphoric
acid is triprotic, the mole ratios are 1:2:1:3 for H3PO4:H2PO−
4:HPO2−
4:PO3−
4.
At the first equivalence point: - Moles of H3PO4= 0.001 mol
At the second equivalence point: - Moles of H3PO4= 0.0025 mol ∇· 2 =
0.00125 mol
At the third equivalence point: - Moles of H3PO4= 0.004 mol ∇· 3 =
0.00133 mol
Step 4: Calculate the concentration of the phosphoric acid solution. - Total
moles of H3PO4= moles at the first equivalence point + moles at the second
equivalence point + moles at the third equivalence point - Total moles of H3PO4
= 0.001 mol + 0.00125 mol + 0.00133 mol = 0.00358 mol
- Concentration of H3PO4= total moles of H3PO4(mol) ∇· volume of
phosphoric acid solution (L) - Concentration of H3PO4= 0.00358 mol ∇· 0.050
L = 0.0716 mol/L
38
