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A weak acid, HA, has a $K_a$ of $1.00 \times 10^{-5}$. If 0.100 mole of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
A
1.00%
B
99.9%
C
0.100%
D
99.0%
Detailed Solution
Given $K_a = 1.00 \times 10^{-5}$ and $C = 0.100$ mol L$^{-1}$
For a weak electrolyte, $HA \rightleftharpoons H^+ + A^-$ with $K_a = \dfrac{C\alpha^2}{1 - \alpha} \approx C\alpha^2$
Degree of dissociation: $\alpha = \sqrt{\dfrac{K_a}{C}}$
$\alpha = \sqrt{\dfrac{1 \times 10^{-5}}{0.100}} = \sqrt{10^{-4}}$
$\alpha = 10^{-2}$
Percentage of acid dissociated $= 10^{-2} \times 100 = 1$ percent
For a weak electrolyte, $HA \rightleftharpoons H^+ + A^-$ with $K_a = \dfrac{C\alpha^2}{1 - \alpha} \approx C\alpha^2$
Degree of dissociation: $\alpha = \sqrt{\dfrac{K_a}{C}}$
$\alpha = \sqrt{\dfrac{1 \times 10^{-5}}{0.100}} = \sqrt{10^{-4}}$
$\alpha = 10^{-2}$
Percentage of acid dissociated $= 10^{-2} \times 100 = 1$ percent