Q. A monoprotic weak acid (HA) is ionised 5% in 0.1 M aqueous solution. What is the equilibrium constant for its ionisation? $HA(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + A^{-}(aq)$
A) $2.63 \times 10^{4}$
B) $2.63 \times 10^{3}$
C) $2.63 \times 10^{- 4}$
D) $2.63 \times 10^{- 3}$
Explanation
(c)
$HA(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + A^{-}(aq)$
Initial 0.1 M 0 0
Change $- \frac{0.1 \times 5}{100}$ $+ \frac{0.1 \times 5}{100}$ $+ \frac{0.1 \times 5}{100}$
Equi conc. 0.095 0.005 0.005
$K_{c} = \frac{\left\lbrack H_{3}O^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack} = \frac{0.005 \times 0.005}{0.095}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.63 \times 10^{- 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.