Question 16.20: Calculate the pH of a 0.10-M solution of sodium fluoride (Na...
Calculate the \text{pH} of a 0.10-M solution of sodium fluoride (\text{NaF}) at 25°\text{C}
Strategy A solution of \text{NaF} contains \text{Na}^+ ions and \text{F}^– ions. The \text{F}^– ion is the conjugate base of the weak acid, \text{HF}. Use the K_a value for \text{HF} (7.1 × 10^{–4} , from Table 16.7) and Equation 16.8 to determine K_b for \text{F}^–:
K_b = \frac{K_w}{K_a} = \frac{1.0 × 10^{–14}}{7.1 × 10^{–4}} = 1.4 × 10^{–11}
Then, solve this \text{pH} problem like any equilibrium problem, using an equilibrium table.
Setup It’s always a good idea to write the equation corresponding to the reaction that takes place along with the equilibrium expression:
\text{F}^–(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{HF}(aq) + \text{OH}^–(aq) K_b = \frac{[\text{HF}][\text{OH}^–]}{[\text{F}^–]}
Construct an equilibrium table, and determine, in terms of the unknown x, the equilibrium concentrations of the species in the equilibrium expression:
\text{F}^–(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{HF}(aq) + \text{OH}^–(aq)
| Initial concentration (M): | 0.10 | 0 | 0 | |
| Change in concentration (M): | –x | +x | +x | |
| Equilibrium concentration (M): | 0.10 − x | x | x |
Equation 16.8 K_a × K_b = K_w
| TA B L E 1 6 . 7 Ionization Constants of Some Weak Acids at 25°C | |||
| Name of acid | Formula | Structure | K_a |
| \text{Chloroacetic acid} | \text{CH}_2\text{ClCOOH} |  |
5.6 × 10^{–2} |
| \text{Hydrofluoric acid} | \text{HF} |  |
7.1 × 10^{–4} |
| \text{Nitrous acid} | \text{HNO}_2 |  |
4.5 × 10^{–4} |
| \text{Formic acid} | \text{HCOOH} |  |
1.7 × 10^{–4} |
| \text{Benzoic acid} | \text{C}_6\text{H}_5\text{COOH} |  |
6.5 × 10^{–5} |
| \text{Acetic acid} | \text{CH}_3\text{COOH} |  |
1.8 × 10^{–5} |
| \text{Hydrocyanic acid} | \text{HCN} |  |
4.9 × 10^{–10} |
| \text{Phenol} | \text{C}_6\text{H}_5\text{OH} |  |
1.3 × 10^{–10} |
Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve for x, we get
1.4 × 10^{–11} = \frac{x^2}{0.10 − x} ≈ \frac{x^2}{0.10}
x = \sqrt{(1.4 × 10^{–11})(0.10)} = 1.2 × 10^{–6} M
According to our equilibrium table, x = [\text{OH}^–]. In this case, the autoionization of water makes a significant contribution to the hydroxide ion concentration so the total concentration will be the sum of 1.2 × 10^{–6} M (from the ionization of \text{F}^– ) and 1.0 × 10^{–7} M (from the autoionization of water)
Therefore, we calculate the \text{pOH} first as
\text{pOH} = –\text{log} (1.2 × 10^{–6} + 1.0 × 10^{–7}) = 5.95
and then the \text{pH},
\text{pH} = 14.00 − \text{pOH} = 14.00 − 5.95 = 8.05
The \text{pH} of a 0.10-M solution of \text{NaF} at 25°\text{C} is 8.05.