Question 15.14: Calculate the pH of: (a) a 0.10 M solution of AlCl3; Ka for ......
Calculate the pH of:
(a) a 0.10 M solution of AlCl_{3}; K_{a} for Al(H_{2}O)_{6} ^{3+} is 1.4\times 10^{-5}.
(b) a 0.10 M solution of NaCN; K_{a} for HCN is 4.9\times 10^{-10}.
STRATEGY
Determine if the salt will be acidic or basic by examining the cations and anions. The pH can be calculated by using the strategy for solving weak acid and base problems outlined in Figure 15.7. Because this problem is similar to others done earlier, we’ll abbreviate the procedure.
(a)
(Steps 1–4. The species present initially are Al(H_{2}O)_{6} ^{3+} (acid), Cl^{-} (inert), and H_{2}O (acid or base). Because Al(H_{2}O)_{6} ^{3+} is a much stronger acid than water (K_{a} >> K_{w}), the principal reaction is dissociation of Al(H_{2}O)_{6} ^{3+}:
Table 1
Step 5. The value of x is obtained from the equilibrium equation:
K_{a}=1.4\times 10^{-5}=\frac{[H_{3}O^{+}][Al(H_{2}O)_{5}(OH)^{2+}]}{[Al(H_{2}O)_{6} ^{3+}]}=\frac{(x)(x)}{(0.10 – x)}\approx \frac{x^{2}}{0.10}x = [H_{3}O^{+}]=1.2\times 10^{-3} M
Step 6. pH = – log (1.2\times 10^{-3}) = 2.92
Thus, Al(H_{2}O)_{6} ^{3+} is a much stronger acid than NH_{4} ^{+}, which agrees with the colors of the indicator in Figure 15.9.
(b)
Step 1. The species present initially are Na^{+} (inert), CN^{-} (base), and H_{2}O (acid or base).
Step 2. There are two possible proton-transfer reactions:
CN^{-}(aq)+H_{2}O(l)\xrightleftharpoons{}HCN(aq)+OH^{-}(aq) K_{b}H_{2}O(l)+H_{2}O(l)\xrightleftharpoons{}H_{3}O^{+}(aq)+OH^{-}(aq) K_{w}
Step 3. K_{b} = K_{w}/(K_{a} for HCN) = 2.0\times 10^{-5}. Because K_{b} >> K_{w}, CN^{-} is a stronger base than H_{2}O and the principal reaction is proton transfer from H_{2}O to CN^{-}.
Step 4.
Table 2
Step 5. The value of x is obtained from the equilibrium equation:
K_{b}=2.0\times 10^{-5}=\frac{[HCN][OH^{-}]}{[CN^{-}]}=\frac{(x)(x)}{(0.10 – x)}\approx \frac{x^{2}}{0.10}x = [OH^{-}]=1.4\times 10^{-3} M
Steps 6–7. [H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{1.4\times 10^{-3}}=7.1\times 10^{-12}
Step 8. pH = – log(7.1\times 10^{-12}) = 11.15
The solution is basic, which agrees with the color of the indicator in Figure 15.9.
Table 1
| Principal reaction | Al(H_{2}O)_{6} ^{3+}(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | H_{3}O^{+}(aq) | Al(H_{2}O)_{5}(OH)^{2+}(aq) |
| Equilibrium concentration (M) | 0.10 – x | x | x |
Table 2
| Principal reaction | CN^{-}(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | HCN(aq) | OH^{-}(aq) |
| Equilibrium concentration (M) | 0.10 – x | x | x |
