Chapter 14: Equilibria in Acid-Base Solutions
2. Refer to Sections 4.2, and 13.4, and Table 13.2.
Write the reactions, eliminating the spectator ions. Bear in mind that soluble salts ionize, thus
they exist in solution as ions. Weak acids and bases (Table 13.2) do not ionize significantly, thus
they exist in solution as the undissociated acid or base.
a. NaC2H3O2(aq) + HNO3(aq) ⇌ HC2H3O2(aq) + NaNO3(aq)
NaC2H3O2 and NaNO3 are salts and HNO3 is a strong acid, thus all three dissociate in
solution, while HC2H3O2 is a weak acid and thus remains undissociated.
Na+(aq) + C2H3O2-(aq) + H+(aq) + NO3-(aq) ⇌ HC2H3O2(aq) + Na+(aq) + NO3-(aq)
C2H3O2-(aq) + H+(aq) ⇌ HC2H3O2(aq)
b. 2HBr(aq) + Sr(OH)2(aq) ⇌ SrBr2(aq) + 2H2O(l)
HBr and Sr(OH)2 are strong acids and bases (respectively) and thus dissociate completely
in solution.
2H+(aq) + 2Br -(aq) + Sr+2(aq) + 2OH-(aq) ⇌ Sr+2 (aq) + 2Br -(aq) + 2H2O(l)
2H+(aq) + 2OH-(aq) ⇌ 2H2O(l)
H+(aq) + OH-(aq) ⇌ H2O(l)
c. HClO(aq) + NaCN(aq) ⇌ NaClO(aq) + HCN(aq)
HClO and HCN are weak acids and do not dissociate; NaCN and NaClO are salts and do
dissociate in solution.
HClO(aq) + Na+(aq) + CN-(aq) ⇌ Na+(aq) + ClO-(aq) + HCN(aq)
HClO(aq) + CN-(aq) ⇌ ClO-(aq) + HCN(aq)
d. HNO2(aq) + NaOH(aq) ⇌ NaNO2(aq) + H2O(l)
HNO2 is a weak acid and does not dissociate; NaOH is a strong base and NaNO2 is a salt,
thus both dissociate in solution.
HNO2(aq) + Na+(aq) + OH-(aq) ⇌ Na+(aq) + NO2-(aq) + H2O(l)
HNO2(aq) + OH-(aq) ⇌ NO2-(aq) + H2O(l)
20. Refer to Section 14.1.
a.
5.50 g x
1 mol. NH 4 Cl
= 0.103 mol. NH 4 Cl
53.49 g NH 4 Cl
[H + ] = K a x
nNH
4
+
nNH 3
+
= 5.6 x 10 -10 x
0.103 mol. NH 4
= 3.1 x 10 - 9 M
0.0188 mol. NH 3
pH = -log[H+] = -log(3.1 x 10-9) = 8.51
b. Note that volume was not used in the calculations above. Thus we would expect the pH
to remain the same, independent of the volume of solution.
42. Refer to Section 14.2, Equation 14.3, and Example 14.5.
a. pKa = -log Ka
Ka = 10-pKa
Ka = 10-7.4 = 4 x 10-8
b. A color change occurs when pH ≈ pKa
[HIn]
[HIn]
The range will be from
≥ 10 to
≤ 0.1
[In ]
[In - ]
Using equation 14.3:
[HIn] [H + ]
=
[In − ] K a
This rearranges to form:
[HIn]
[HIn]
⇒ - log [H + ] = - log K a - log
−
[In ]
[In − ]
[HIn]
pH = p K a - log
= 7.4 - log 10 = 6.4
[In − ]
[HIn]
pH = p K a - log
= 7.4 - log 0.1 = 8.4
[In − ]
[H + ] = K a x
Thus the range is from pH 6.4 to 8.4.
c.
When the pH = pKa = 7.4 one would be half way between red and yellow, thus the color
would be orange.