Question 14.16: CALCULATING THE pH OF A BASIC SALT SOLUTION Calculate the pH...
CALCULATING THE pH OF A BASIC SALT SOLUTION
Calculate the pH of a 0.10 M solution of NaCN; K_{a} for HCN is 4.9 × 10^{-10}.
STRATEGY
Use the procedure summarized in Figure 14.6.
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) \rightleftharpoons HCN(aq) + OH^{-}(aq) K_{b}
H_{2}O(l) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + OH^{-}(aq) K_{w}
Step 3. As shown in Worked Example 14.14b, K_{b} = K_{w}/(K_{a} for HCN) = 2.0 × 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.
Step 5. The value of x is obtained from the equilibrium equation:
K_{b} = 2.0 × 10^{-5} = \frac{[HCN][OH^{-}]}{[CN^{-}]} = \frac{(x)(x)}{(0.10 – x)} ≈ \frac{x^{2}}{0.10}x = [OH^{-}] = 1.4 × 10^{-3} M
Step 7. [H_{3}O^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{1.0 × 10^{-14}}{1.4 × 10^{-3}} = 7.1 × 10^{-12}
Step 8. pH = -log (7.1 × 10^{-12}) = 11.15
The solution is basic, which agrees with the color of the indicator in Figure 14.8.
| Principal reaction CN^{-}(aq) + H_{2}O(l) \rightleftharpoons HCN(aq) + OH^{-}(aq) |
| Equilibrium concentration (M) 0.10 – x x x |
