Question 16.13: The Ka of hypochlorous acid (HClO) is 3.5 × 10^–8 . Calculat...
The K_a of hypochlorous acid (\text{HClO}) is 3.5 × 10^{–8} . Calculate the \text{pH} of a solution at 25°\text{C} that is 0.0075 M in \text{HClO}.
Strategy Construct an equilibrium table, and express the equilibrium concentration of each species in terms of x. Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid. Use Equation 16.2 to determine \text{pH}.
Setup
\text{HClO}(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{H}_3\text{O}^+(aq) + \text{ClO}^–(aq)
| Initial concentration (M): | 0.0075 | 0 | 0 | |
| Change in concentration (M): | –x | +x | +x | |
| Equilibrium concentration (M): | 0.0075 − x | x | x |
\text{pH} = –\text{log} [\text{H}_3\text{O}^+] or pH = –\text{log} [\text{H}^+] Equation 16.2
These equilibrium concentrations are then substituted into the equilibrium expression to give
K_a = \frac{(x)(x)}{0.0075 − x} = 3.5 × 10^{–8}
Assuming that 0.0075 − x ≈ 0.0075,
\frac{x ^2}{0.0075} = 3.5 × 10^{–8} x^2 = (3.5 × 10^{–8})(0.0075)
Solving for x, we get
x = \sqrt{2.625 × 10^{–10}} = 1.62 × 10^{–5} M^*
According to the equilibrium table, x = [\text{H}_3\text{O}^+]. Therefore,
\text{pH} = –\text{log} (1.62 × 10^{–5}) = 4.79
^*Student Annotation: Applying the 5 percent test indicates that the approximation shortcut is valid in this case: (1.62 × 10^{–5} /0.0075) × 100\% < 5\%.