Question 14.15: CALCULATING THE pH OF AN ACIDIC SALT SOLUTION Calculate the ...
CALCULATING THE pH OF AN ACIDIC SALT SOLUTION
Calculate the pH of a 0.10 M solution of AlCl_{3}; K_{a} for Al(H_{2}O)_{6}^{3+} is 1.4 × 10^{-5}.
STRATEGY
Because this problem is similar to others done earlier, we’ll abbreviate the procedure in Figure 14.6.
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+}:
Step 5. The value of x is obtained from the equilibrium equation:
K_{a} = 1.4 × 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)} ≈ \frac{x^{2}}{0.10}x = [H_{3}O^{+}] = 1.2 × 10^{-3} M
Step 8. pH = -log (1.2 × 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 14.8.
| Principal reactio Al(H_{2}O)_{6}^{3+}(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + Al(H_{2}O)_{5}(OH)^{2+}(aq) |
| Equilibrium 0.10 – x x x
concentration (M) |
