LIST OF TOPICS TEST – ACID/BASE EQUILIBRIUM
Be able to calculate the pH of any solution of strong base, strong acid, weak base, weak acid or buffer
Know the differences between a weak acid and a strong acid (same for bases)
Know that at the pKa the [HA] equals the [A-]
Know that a buffer is a solution that is a 1:1 ratio of a weak acid with its conjugate base and will have a pH equal to
the pKa of the weak acid.
Know that a buffer is made by adding NaOH drop by drop to a solution of a weak acid until the pH reaches the pKa
Know that the buffer you want to use will have a pKa equal to the pH at which you want to buffer (#85)
Know that an indicator is a weak acid.
Know that an indicator is one color in the acid form (HIn) and another color in the base form (IN-)
Know that at the pKa since [HA] equals the [A-] the color will change at the pKa
Know that when doing a titration you choose an indicator that will change color at the equivalence (stoichiometric)
point.
Indicator chap 15 #69, 79
Be able to calculate any unknown out of the buffer equation: pKa = pH – log ([A-]/[HA])
Know all the parts of any titration curve (chap 15 #55 is a good example for one of the curves)
Know how to calculate the pH at any point in a titration.
Particularly know how to calculate the pH in a weak acid/strong base titration. (chap 15 #59 is a good example)
Know that Ka reflects acid strength and that pKa is found in the titration curve. (ch 15 #90)
Chap 15 # 83 ab, 85, 91, 93, 35, 37
Free response solutions
1970
(a) What is the pH of a 2.0 molar solution of acetic acid? Ka acetic acid = 1.810–5
(b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter
of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the
buffer solution.
(c) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer
prepared in (b). Compute the hydrogen ion concentration of the resulting solution.
Answer:
(a) CH3COOH(aq) ↔ H+(aq) + CH3COO–(aq)
Ka = 1.810-5 =
[H  ][CH3COO ]
[CH3COOH]
[H+] = [CH3COO–] = X
[CH3COOH] = 2.0 – X, X << 2.0, (2.0– X) = 2.0
X2
2.0
–3
X = 6.010 = [H+]; pH = –log [H+] = 2.22
(b) 0.1 L  2.0 mol/L = 0.20 mol CH3COOH
0.1 L  1.0 mol/L = 0.10 mol NaOH
the 0.10 mol of hydroxide neutralizes 0.10 mol CH3COOH with 0.10 mol remaining with a
concentration of 0.10 mol/0.20 L = 0.5 M. This also produces 0.10 mol of acetate ion in 0.20
L, therefore, [CH3COO–] = 0.50 M.
1.810–5 =
[H  ][0.50]
1.810 =
(0.50)
+
[H ] = 1.810–5 = pH of 4.74
(c) [CH3COOH]o = [CH3COO–]o
0.040 L
= 0.50 M 
= 0.143 M
0.14 L
-5
[H+]o = 0.50 M 
0. 10 L
0. 14 L
= 0.357 M
the equilibrium will be pushed nearly totally to the left resulting in a decrease of the
hydrogen ion by 0.143M. Therefore, the [H+]aq = 0.357M – 0.143M = 0.214M.
2007 part A, question #1
HF(aq) + H2O(l)  H3O+(aq) + F–(aq) Ka = 7.2×10–4
Hydrofluoric acid, HF(aq), dissociates in water as represented by the equation above.
(a) Write the equilibrium-constant expression for the dissociation of HF(aq) in water.
(b) Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.
HF(aq) reacts with NaOH(aq) according to the reaction represented below.
HF(aq) + OH–(aq)  H2O(l) + F–(aq)
A volume of 15 mL of 0.40 M NaOH(aq) is added to 25 mL of 0.40 M HF(aq) solution. Assume
that volumes are additive.
(c) Calculate the number of moles of HF(aq) remaining in the solution.
(d) Calculate the molar concentration of F–(aq) in the solution.
(e) Calculate the pH of the solution.
Answer:
[H3O+ ][F- ]
[HF]
(b) let X = [H3O+] = [F–]; let (0.40-X) = [HF]
(X)(X)
7.2×10–4 =
; X = 1.7×10-2 = [H3O+]
(0.40-X)
(c) (0.40 M)(15 mL) = 6.0 mmol OH– neutralizes and equal number of moles of HF
(0.40 M)(25 mL) = 10. mmol HF
(a) K a 
4.0 mmol HF remains (or 0.0040 mol)
(d) 6.0 mmol OH produces 6.0 mmol of F–, from a 1:1 stoichiometry in the equation
–
6.0 mmol Fnow in 40. mL of solution, [F ] =
= 0.15 M
40. mL
( X )(0.15)
4.0 mmol HF
(e)
= 0.10 M HF; let X = [H3O+]; 7.2×10–4 =
; X = 4.8×10–4
40. mL
(0.10  X )
+
–4
pH = –log[H3O ] = –log(4.8×10 ) = 3.32
–
2005 A Required
HC3H5O2(aq) ↔ C3H5O2–(aq) + H+(aq)
Ka = 1.3410–5
Propanoic acid, HC3H5O2, ionizes in water according to the equation above.
(a) Write the equilibrium constant expression for the reaction.
(b) Calculate the pH of a 0.265 M solution of propanoic acid.
(c) A 0.496 g sample of sodium propanoate, NaC3H5O2, is added to a 50.0 mL sample of a
0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution
occurs, calculate each of the following.
(i) The concentration of the propanoate ion, C3H5O2–(aq) in the solution
(ii) The concentration of the H+(aq) ion in the solution.
The methanoate ion, HCO2–(aq) reacts with water to form methanoic acid and hydroxide ion, as
shown in the following equation.
HCO2–(aq) + H2O (l) ↔ H2CO2(aq) + OH–(aq)
(d) Given that [OH–] is 4.1810–6 M in a 0.309 M solution of sodium methanoate, calculate
each of the following.
(i) The value of Kb for the methanoate ion, HCO2–(aq)
(ii) The value of Ka for methanoic acid, HCO2H
(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
Answer:
[C3H5O-2 ][H + ]
(a)
= Ka
[HC3H5O 2 ]
(b) let X be the amount of acid that ionizes, then
X = [C3H5O2–] = [H+]
0.265 – X = [HC3H5O2]
X2
= Ka = 1.3410–5
0.265 X
X= 0.00188 M = [H+]
[you can assume that 0.265 – X ≈ 0.265 in order to simplify your calculations]
pH = –log[H+] = 2.73
2003 A Required
C6H5NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH–(aq)
Aniline, a weak base, reacts with water according to the reaction represented above.
(a) Write the equilibrium constant expression, Kb, for the reaction represented above.
(b) A sample of aniline is dissolved in water to produce 25.0 mL of 0.10 M solution. The pH of
the solution is 8.82. Calculate the equilibrium constant, Kb, for this reaction.
(c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calculate the pH of the
solution when 5.0 mL of the acid has been titrated.
(d) Calculate the pH at the equivalence point of the titration in part (c).
(e) The pKa values for several indicators are given below. Which of the indicators listed is most
suitable for this titration? Justify your answer.
Indicator
pKa
Erythrosine
3
Litmus
7
Thymolphthalein 10
Answer:
[C6 H5 NH 3 ][OH  ]
(a) Kb =
[C6 H5 NH 2 ]
(b) pOH = 14 – pH = 14 – 8.82 = 5.18
–log[OH–] = 5.18; [OH–] = 6.6110–6 M
[OH–] = [C6H5NH3+]
 6.6110 
6 2
Kb =
6
= 4.410–10
0.10 – 6.61 10
0.1 mol
(c) 25 mL 
= 2.5 mmol C6H5NH2
1L
0.1 mol
5 mL 
= 0.5 mmol H+ added
1L
2.0 mmol base remains in 30.0 mL solution
0.50 mmol 
 X   X 
30.0 mL 

4.410–10 =
 20.0 mmol 
 30.0 mL 
X = 1.8010–9 = [OH–]
11014
[H ] =
= 5.610–6; pH = 5.26
9
1.8 10
(d) when neutralized, there are 2.5 mmol of C6H5NH3+ in 50.0 mL of solution, giving a
[C6H5NH3+] = 0.050 M
this cation will partially ionize according to the following equilibrium:
C6H5NH3+(aq) ↔ C6H5NH2(aq) + H+(aq)
+
at equilibrium, [C6H5NH2] = [H+] = X
[C6H5NH3+] = (0.050–X)
X2
= Ka = 2.310–5
(0.050  X )
X = 1.0610–3 = [H+]
pH = –log[H+] = 2.98
(e) erythrosine; the indicator will change color when the pH is near its pKa, since the
equivalence point is near pH 3, the indicator must have a pKa near this value.