Question 16.13: The Ka of hypochlorous acid (HClO) is 3.5 × 10^–8 . Calculat...

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: $(1.62 × 10^{–5} /0.0075) × 100\% < 5\%$.