Q. By applying law of mass action, the equilibrium constant, $K$ for the reaction $HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}$, is given as
A) $K = \frac{\lbrack HA\rbrack\lbrack H_{2}O\rbrack}{\left\lbrack H_{3}O^{+} \right\rbrack\lbrack A^{-}\rbrack}$
B) $K = \frac{\left\lbrack H_{3}O^{+} \right\rbrack\lbrack A^{-}\rbrack}{\lbrack HA\rbrack\lbrack H_{2}O\rbrack}$
C) $K = \frac{\left\lbrack H_{3}O^{+} \right\rbrack\lbrack H_{2}O\rbrack}{\left\lbrack A^{-} \right\rbrack\lbrack HA\rbrack}$
D) $K = \frac{\lbrack HA\rbrack\lbrack A^{-}\rbrack}{\left\lbrack H_{2}O \right\rbrack\lbrack H_{3}O^{+}\rbrack}$
Explanation
(b)
$HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}$
$K = \frac{\left\lbrack H_{3}O^{+} \right\rbrack\lbrack A^{-}\rbrack}{\lbrack HA\rbrack\lbrack H_{2}O\rbrack}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.