South Pasadena • AP Chemistry
Name
4 ▪ Acid Base Solutions
Period
4.3
PROBLEMS
1. Using the list of Ka/Kb, arrange the following
substances in order of increasing acid strength.
H2O, H2SO3, HCN, H2PO4−, NH4+,
HCO2H, HCl.
H2O
Ka = 1.0 × 10−14
pKa = 14.00
+
−10
NH4
Ka = 5.6 × 10
pKa = 9.26
−10
HCN
Ka = 6.2 × 10
pKa = 9.21
−
−8
H2PO4
Ka = 6.2 × 10
pKa = 7.21
−4
HCOOH Ka = 1.8 × 10
pKa = 3.74
−2
H2SO3
Ka = 1.5 ×10
pKa = 1.82
7
HCl
Ka ≈ 10
pKa = −7
2. Given the following values:
 Ka2 for H2SO4 is 1.2 × 10−2
 Ka for HCN is 6.2 × 10−10
 Ka for HF is 7.2 × 10−4
 Kas for H2CO3 are 4.3 × 10−7 and 5.6 × 10−11
 Kb for NH3 is 1.8 × 10−5
Determine among these acids: HNO3, HSO4−,
HCN, H2CO3, NH4+, HF
a. The strongest weak acid
HSO4−
b. The acid that produces the lowest [H3O+] per
mole of acid
NH4+ (lowest Ka: 5.5 × 10−10)
c. The acid with the strongest conjugate base
NH4+ (weakest conjugate acid)
d. The diprotic acid
H2CO3
e. The strong acid
HNO3
f.
The acid with the weakest conjugate base
HNO3
–
Date
STRENGTH OF
ACIDS
3. Rank the following salts, when dissolved in water
to produce 0.10 M solutions, from highest to
lowest pH.
NaC2H3O2, Zn(NO3)2, KCl, KCN, NaHSO4
KCN
Kb of CN− = 1.6 × 10−5
NaC2H3O2
Kb of C2H3O2− = 5.6 × 10−10
KCl
KCl is neutral
NaHSO4
Ka of HSO4− = 1.2 × 10−2
Zn(NO3)2
Zn2+ is acidic
4. Consider the bicarbonate ion, HCO3−. This ion is
amphiprotic: it can both donate and accept an H+.
However, baking soda, NaHCO3, forms a basic
solution.
a. Write the acid dissociation equation (Ka) for
HCO3− and look up its value. (This is the K2
for H2CO3.)
HCO3− + H2O  H3O+ + CO32−
[H3O+][CO32−]
K2 =
= 5.6 × 10−11
[HCO3−]
b. Write the base dissociation equation (Kb) for
HCO3− and calculate its value. (Use the K1 for
H2CO3.)
HCO3− + H2O  H2CO3 + OH−
[H2CO3][OH−] Kw 1.0 × 10−14
Kb =
=
=
[HCO3−]
Ka1 4.3 × 10−7
= 2.3 × 10−8
c. Consider the two reactions. Explain why
NaHCO3 would form a basic solution based on
your calculations.
NaHCO3 is a basic solution. Na+ is neutral.
HCO3− is basic because its Kb > Ka.
5. Cyanic acid HOCN has a Ka = 3.5 × 10−4, what is
the Kb for the cyanate ion OCN−?
Kw 1.0 × 10−14
Kb =
=
= 2.9 × 10−11
Ka
3.5 × 10−4
6. Phenol is a relatively weak acid, Ka = 1.3 × 10−10.
How does the strength of its conjugate base
compare with the strength of ammonia, the acetate
ion, and sodium hydroxide? (Use the list of
Ka/Kb.)
C2H3O2−
Kb = 5.6 × 10−10
NH3
Kb = 1.8 ×10−5
Conjugate base of Phenol Kb = 7.7 × 10−5
OH−
Kb = very large
8. Consider a 0.80 M solution of H2SO3.
K1 = 1.5 × 10−2, K2 = 1.0 × 10−7.
a. What are the pH and concentration of all
species?
[H+][HSO3−]
K1 =
= 1.5 × 10−2
[H2SO3]
(x)(x)
= 1.5 × 10−2
(0.80 − x)
(Use quadratic formula. Cannot approx.
x << 0.80 because K1 is too big.)
x = [H+] = [HSO3−] = 0.100 M
pH = −log(0.100) = 1.000
[H2SO3] = 0.80 – 0.10 = 0.70 M
[H+][SO32−]
= 1.0 × 10−7
[HSO3−]
(0.100 + x)(x)
= 1.0 × 10−7
(0.100 − x)
K2 =
7. Consider the following processes. Do they favor
the reactants or products?
a. NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
Reactant Favored
(H2O is weaker than NH4+)
b. HCN(aq) + H2O(l)  H3O+(aq) + CN−(aq)
Reactant Favored
(HCN is weaker than H3O+)
c. NH4 (aq) + CO3 (aq)  NH3(aq) +HCO3 (aq)
Product Favored
(NH4+ is stronger than HCO3−)
+
2−
−
Assume x << 0.100
x = [SO32−] = 1.0 × 10−7 M
b. What happens to the [SO32−] if the
concentration of sulfurous acid is halved?
[SO32−] does not change.
9. Rank the following acids from weakest to
strongest. Explain your reason for each one.
a. HClO2, HBrO2, and HIO2
HIO2 < HBrO2 < HClO2
Cl is most electronegative, so polarizes
electrons away from H−O bond most,
resulting in the weakest H−O bond and
strongest acid.
b. H2O, H2S, H2Se
H2O < H2S < H2Se
Se has the largest ionic radius, so it has the
weakest H−X bond, resulting in being the
strongest acid.
c. HIO, HIO2, HIO3
HIO < HIO2 < HIO3
HIO3 has the most oxygens, so electrons are
polarized away from H−O bond most,
resulting in the weakest H−O bond and
strongest acid.
AP Chemistry 2009 #1
Answer the following questions that relate to the chemistry of halogen oxoacids.
(a) Use the information in 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 K a.
HOCl is the stronger acid because it has a larger Ka value.
(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.
HOI is a weaker acid than HOCl because I is less electronegative than Cl, so the H−O bond in HOI
is stronger than that of HOCl.
(b) Write the equation for the reaction that occurs between hypochlorous acid and water.
HOCl + H2O → OCl− + H3O+
(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).
[HOCl][OH−]
Kb =
[OCl−]
(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.
Kw 1.0 × 10−14
Kb =
=
= 3.4 × 10−7
Ka 2.9 × 10−8
(iii) Calculate the value of [OH–] in the 1.2 M NaOCl solution at 298 K .
[HOCl][OH−]
(x)(x)
Kb =
=
= 3.4 × 10−7
[OCl−]
(1.2 − x)
x = [OH−] = 6.4 × 10−4 M
(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.
[H3O+] = 10−pH = 10−6.48 = 3.3 × 10−7
(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.
[OCl−]
pH = pKa + log
[HOCl]
[OCl−]
[OCl−]
6.48 = −log (2.9 × 10−8) + log
= 7.54 + log
[HOCl]
[HOCl]
−
−
[OCl ]
[OCl ]
log
= −1.05
Because log
< 0, so [OCl−] < [HOCl] at this buffer.
[HOCl]
[HOCl]
AP Chemistry 1990 #8
Give a brief explanation for each of the following.
(a) For the diprotic acid H2S, the first dissociation constant is larger than the second dissociation constant by
about 105 (K1 = 105 K2).
The second dissociation constant removes an H+ from the charged HS− ion, which requires more energy
than removing the first H+ from the neutral H2S molecule.
(b) In water, NaOH is a base, but HOCl is an acid.
NaOH is a soluble salt, which dissociates into the Na+ and OH− ions, resulting in a basic solution.
HOCl is an acid because it has a weak H−O bond. It dissociates into water to H+ + OCl−.
(c) HCl and HI are equally strong acids in water but, in pure acetic acid, HI is a stronger acid than HCl.
Water is a strong enough H+ acceptor to mask any difference in donor strength of these two strong
acids. Acetic acid is a very weak H+ acceptor and can show that HI is a better donor.
(d) When each is dissolved in water, HCl is a much stronger acid than HF.
The Cl− ion is larger than F−, so the H−Cl bond is weaker than the H−F bond, resulting in Cl being a
stronger acid than HF.