- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
hard
A weak acid, $HA,$ has a $K_a$ of $1.00 \times 10^{-5}.$ If $0.100 \,mol$ of this acid is dissolved in one litreof water, the percentage of acid dissociated at equilibrium is closest to.....$\%$
A
$1$
B
$99.9$
C
$0.1$
D
$99$
(AIPMT-2007)
Solution
$H A \rightleftharpoons H^{+}+A$
At equilibrium $\left[H^{+}=A^{-}\right]$
$K_{a} =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}=\frac{\left[H^{+}\right]^{2}}{[H A]}$
$\left[H^{+}\right]=\sqrt{K_{a}[H A]} =\sqrt{1 \times 10^{-5} \times 0.1}$
$= \sqrt{1 \times 10^{-6}}=1 \times 10^{-3}$
$\alpha =\frac{A \text {ctual ionisation}}{\text {Molar concentration}}$
$\%$ of acid dissociated $=10^{-2} \times 1.00$
$=1 \%$
Standard 11
Chemistry
