AP Acid-Base Problems
Problem # 1
HOBr(aq) ↔ H+(aq) + OBr–(aq)
Ka = 2.3  10–9
Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation above.
(a) Calculate the value of [H+] in an HOBr solution that has a pH of 4.95.
(b) Write the equilibrium constant expression for the ionization of HOBr in water, then calculate the
concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.8  10–5 M.
(c) A solution of Ba(OH)2 is titrated into a solution of HOBr.
(i) Calculate the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point when titrated into
a 65.0 mL sample of 0.146 M HOBr(aq).
(ii) Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7. Explain.
(d) Calculate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160 M HOBr to
produce a buffer solution with [H+] = 5.00  10–9 M. Assume that volume change is negligible.
(e) HOBr is a weaker acid than HBrO3. Account for this fact in terms of molecular structure.
Answer:
(a) pH = –log[H+]; 4.95 = –log[H+]
[H+] = 1.12  10–5
[H + ][OBr - ]
(b) Ka =
= 2.310–9
[HOBr]
[H+] = [OBr–] = 1.810–5 M
(1.8  10 5 )2
= 2.310–9
X
X = [HOBr] = 0.14 M
1000 mL
0.146 mol HOBr
1 mol H +
1 mol OH - 1 mol Ba(OH)2
(c) (i) 65.0 mL 




=
+
1000 mL
2 mol OH
0.115 mol Ba(OH)2
1 mol HOBr
1 mol H
= 41.3 mL
(ii) pH > 7; salt of a weak acid is a weak base
[A]
[HA]K a
(d) pH = pKa + log
; [A–] =
[HA]
[H  ]
(0.160)(2.3  10 9 )
=
= 0.0736 M
5.00  10 9
0.736 mol OBr - 1 mol NaOBr
125 mL 

= 0.00920 mol NaOBr
1 mol OBr 1000 mL
(e) very electronegative oxygen is able to draw electrons away from the bromine and weaken the O–H bond,
making it easier for the hydrogen ion “to leave”.
[OBr–]
Problem # 2
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:
Problem # 3
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:
[C3H 5O-2 ][H + ]
(a)
= Ka
[HC3H 5O2 ]
(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
1 mol
96.0 g
(c) (i)
= 0.103 M
0.050 L
since each sodium propanoate dissociates completely when dissolved, producing 1 propanoate ion for every
sodium propanoate, and this is over 1000’s of times larger than the propanoate ions from the acid, then
[C3H5O2–] = 0.103 M
(ii) let X be the amount that ionizes, then:
X = [H+]
X + 0.103 = [C3H5O2–]
0.265 – X = [HC3H5O2]
(X)(0.103  X)
= Ka = 1.3410–5
0.265  X
X= 3.4310–5 M = [H+]
[you can assume that 0.265 – X ≈ 0.265 and X + 0.103 ≈ 0.103, in order to simplify your calculations]
(d) (i) [HCO2] = [OH–] = 4.1810–6 M
[HCO2–] = 0.309 – 4.1810–6
0.496 g 
[H 2CO2 ][OH- ]
(4.18  10 6 )2
Kb =
=
= 5.6510–11
[HCO2 - ]
(0.309  4.18  10 6 )
(ii) Ka =
Kw
1  1014
=
= 1.7710–4
Kb
5.65  1011
(e) methanoic acid is stronger;
the larger the Ka, the stronger the acid
OR
for monoprotic organic acids, the longer the carbon chain, the weaker the acid. Propanoic has 3 carbons,
whereas, methanoic has only 1.
Problem # 4
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:
[H O+ ][F - ]
(a) K a  3
[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
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 F –
now in 40. mL of solution, [F ] =
= 0.15 M
40. mL
4.0 mmol HF
(X)(0.15)
(e)
= 0.10 M HF; let X = [H3O+]; 7.2×10–4 =
; X = 4.8×10–4
40. mL
(0.10  X)
pH = –log[H3O+] = –log(4.8×10–4) = 3.32
Problem # 5
Answer the following questions that relate to the chemistry of halogen oxoacids.
(a) Use the information the table below to answer part (a)(i).
Acid
Ka at 298 K
HOCl
2.9  10-8
HOBr
2.4  10-9
(i) Which of the two acids is stronger, HOCl or HOBr? Justify your answer in terms of Ka.
(ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part (a)(i).
(iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than
the acid that you identified in part (a)(i). Justify your prediction in terms of chemical bonding.
(b) Write the equation for the reaction that occurs between hypochlorous acid and water.
(c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The hydrolysis
reaction OCl-(aq) + H2O(l)  HOCl(aq) + OH-(aq) occurs.
(i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs between OCl-(aq) and
H2O(l).
(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.
(iii) Calculate the value of [OH-] in the 1.2 M NaOCl solution at 298 K.
(d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 298 K. The pH of the
buffer solution is determined to be 6.48.
(i) Calculate the value of [H3O+] in the buffer solution.
(ii) Indicate which of HOCl(aq) or OCl-(aq) is present at the higher concentration in the buffer solution.
Support your answer with a calculation.
Answer:
(a) (i) HOCl, larger Ka means higher percentage of ionization and, therefore, a stronger acid
gg
gg
gg
gg
(ii) H-O Cl :
(iii) weaker, the H-O bond strength is stronger when the other element bonded to the oxygen has a lower
electronegativity (iodine has a lower EN than chlorine). Thus a H-O-I molecule holds on to its H more
strongly than HOCl and less H+ ionizes
(b) HOCl + H2O  OCl– + H3O+
[HOCl][OH- ]
Kw
1  10 14

K


(c) (i) & (ii)
= 3.4  10-7
b
[OCl– ]
K a 2.9  10 8
(iii) [HOCl] = [OH-] = X
[OCl-] = 1.2 - X
X2
= 3.4  10-7 ; X = 6.4  10-4 M = [OH-]
1.2  X
Problem # 6
1. A pure 14.85 g sample of the weak base ethylamine, C2H5NH2 , is dissolved in enough distilled water to
make 500. mL of solution.
(a) Calculate the molar concentration of the C2H5NH2 in the solution.
The aqueous ethylamine reacts with water according to the equation below.
C2H5NH2(aq) + H2O(l)  C2H5NH3+(aq) + OH-(aq)
(b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water.
(c) Of C2H5NH2(aq) and C2H5NH3+(aq) , which is present in the solution at the higher concentration at
equilibrium? Justify your answer.
(d) A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl.
Assume that volumes are additive. The pH of the resulting solution is found to be 10.93.
(i) Calculate the concentration of OH-(aq) in the solution.
(ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is
mixed with the HCl solution.
(iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction.
(iv) Calculate the value of Kb for C2H5NH2 .
Answer:
1 mol
14.85g 
moles
45.0 g
(a) molarity =
=
= 0.660 M
liter
0.500 L
[C H NH +3 ][OH - ]
(b) Kb = 2 5
[C2 H 5 NH 2 ]
(c) C2H5NH2; weak base means Kb is very small so little of the ethylamine dissociates
(d) (i) pOH = 14 – pH = 14 – 10.93 = 3.07
pOH = -log[OH–] = 3.07; [OH–] = 8.5110-4 M
(ii) C2H5NH2(aq) + H+(aq)  C2H5NH3+(aq)
(iii) (0.500 L)(0.200 M) = 0.100 mol H+ = # moles of C2H5NH3+ since the stoichiometry is 1:1
0.100 mol C2H5NH3+ /1 L of solution = 0.100 M
(iv) [C2H5NH3+] = 0.100 M; [OH–] = 8.510-4
Kb =
(0.100)(8.51  10 4 )
= 5.6710-4
(0.150)