Calculate the $pH$ of a $0.08 \ M$ solution of hypochlorous acid,$HOCl$. The ionization constant of the acid is $2.5 \times 10^{-5}$. Determine the percent dissociation of $HOCl$.
$HOCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + ClO^{-}_{(aq)}$
Initial concentration $(M)$:
$0.08 \quad \quad \quad \quad \quad \quad \quad 0 \quad \quad \quad \quad 0$
Equilibrium concentration $(M)$:
$(0.08 - x) \quad \quad \quad \quad x \quad \quad \quad \quad x$
$K_a = \frac{[H_3O^{+}][ClO^{-}]}{[HOCl]} = \frac{x^2}{0.08 - x} \approx \frac{x^2}{0.08} = 2.5 \times 10^{-5}$
$x^2 = 2.0 \times 10^{-6} \implies x = 1.41 \times 10^{-3} \ M$
$[H^{+}] = 1.41 \times 10^{-3} \ M$
$pH = -\log(1.41 \times 10^{-3}) = 2.85$
Percent dissociation $= \frac{[HOCl]_{dissociated}}{[HOCl]_{initial}} \times 100 = \frac{1.41 \times 10^{-3}}{0.08} \times 100 = 1.76 \%$
Initial concentration $(M)$:
$0.08 \quad \quad \quad \quad \quad \quad \quad 0 \quad \quad \quad \quad 0$
Equilibrium concentration $(M)$:
$(0.08 - x) \quad \quad \quad \quad x \quad \quad \quad \quad x$
$K_a = \frac{[H_3O^{+}][ClO^{-}]}{[HOCl]} = \frac{x^2}{0.08 - x} \approx \frac{x^2}{0.08} = 2.5 \times 10^{-5}$
$x^2 = 2.0 \times 10^{-6} \implies x = 1.41 \times 10^{-3} \ M$
$[H^{+}] = 1.41 \times 10^{-3} \ M$
$pH = -\log(1.41 \times 10^{-3}) = 2.85$
Percent dissociation $= \frac{[HOCl]_{dissociated}}{[HOCl]_{initial}} \times 100 = \frac{1.41 \times 10^{-3}}{0.08} \times 100 = 1.76 \%$
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