Question 16.2: Calculate the concentrations of all species present, the pH,......
Calculate the concentrations of all species present, the pH, and the percent dissociation of ammonia in a solution that is 0.15 M in NH_{3} (K_{b} = 1.8\times 10^{-5}) and 0.45 M in NH_{4}Cl. (For comparison, the pH of a 0.15 M NH_{3} solution that contains no NH_{4}Cl is 11.21 and the percent dissociation is 1.1%.)
STRATEGY
The solution contains a weak base (NH_{3}) and its conjugate acid (NH_{4} ^{+}), the common ion. Because this problem is similar to the acetic acid–sodium acetate problem, we’ll abbreviate the procedure shown in Figure 15.7.
| IDENTIFY | |
| Known | Unknown |
| Solution concentration (0.15 M NH_{3} and 0.45 M NH_{4}Cl) | pH |
| For NH_{3}, K_{b}=1.8\times 10^{-5} | Concentration of all species Percent dissociation |
Steps 1–3. The principal reaction is proton transfer to NH_{3} from H_{2}O:
NH_{3}(aq)+H_{2}O(l)\xrightleftharpoons{}NH_{4} ^{+}(aq)+OH^{-}(aq)Step 4. Since NH_{4} ^{+} ions come both from the NH_{4}Cl present initially (0.45 M) and from the reaction of NH_{3} with H_{2}O, the concentrations of the species involved in the principal reaction are as follows:
Table 1
Steps 5–6. The equilibrium equation for the principal reaction is
K_{b}=1.8\times 10^{-5}=\frac{[NH_{4} ^{+}][OH^{-}]}{[NH_{3}]}=\frac{(0.45 + x)(x)}{0.15 – x}\approx \frac{(0.45)(x)}{0.15}We assume x is negligible compared to 0.45 and 0.15 because (1) the equilibrium constant K_{b} is small and (2) the equilibrium is shifted to the left by the common-ion effect. Therefore,
x = [OH^{-}]=\frac{(1.8\times 10^{-5})(0.15)}{0.45}=6.0\times 10^{-6} M[NH_{3}]=0.15 – x =0.15-(6.0\times 10^{-6})=0.15 M
[NH_{4} ^{+}]=0.45 + x =0.45+(6.0\times 10^{-6})=0.45 M
Thus, the assumption concerning the negligible size of x is justified.
Steps 7–8. The H_{3}O^{+} concentration and the pH are
[H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{6.0\times 10^{-6}}=1.7\times 10^{-9} MpH =-log (1.7\times 10^{-9})=8.77
The percent dissociation of ammonia is
Percent dissociation =\frac{[NH_{3}]_{dissociated}}{[NH_{3}]_{initial}}\times 100\%=\frac{6.0\times 10^{-6}}{0.15}\times 100\%
= 0.0040 %
Table 1
| pal reaction | NH_{3}(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | NH_{4} ^{+}(aq) | + | OH^{-}(aq) |
| Initial concentration (M) | 0.15 | 0.45 | ∼ 0 | ||||
| Change (M) | -x | +x | +x | ||||
| Equilibrium concentration (M) | 0.15 – x | 0.45 + x | x |