A weak acid, HA and a $\mathrm{K}_a$ of $1.00 \times 10^{-5}$. If $0.100 \mathrm{~mol}$ of this acid is…
A weak acid, HA and a $\mathrm{K}_a$ of $1.00 \times 10^{-5}$. If $0.100 \mathrm{~mol}$ of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closer to:
$1.00 \%$
$99.9 \%$
$0.100 \%$
$99.0 \%$
Solution
Given $\mathrm{K}_a=1.00 \times 10^{-5}, \mathrm{C}=0.100$ mol
for a weak electrolyte, degree of dissociation
$\propto=\sqrt{\frac{\mathrm{K}_a}{\mathrm{C}}}=\sqrt{\frac{1 \times 10^{-5}}{0.100}}=10^{-2}=1 \%$
Related Theory
In any acid-base reaction, the equilibrium will favour the reaction that moves the proton to the stronger base. This equilibrium constant is referred to as the ion-product constant for water, Kw. In pure water, some molecules act as bases and some as acids.