Question

The value of equilibrium constant for self ionisation of water at ${25}^{\circ }C$ is (density of water 1 $g/ml$) :

• A. $55.5\times {10}^{-18}$
• B. $5.55\times {10}^{-18}$
• C. $3.25\times {10}^{-18}$
• D. $1\times {10}^{-18}$

Answer 1

The correct option is C $3.25\times {10}^{-18}$

The self ionization reaction for water is:
$H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$
Therefore equilibrium constant ($K$),
$K=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{[H_{2}O{]}^2}$
The density of water is $1g/mL$.
So, $\left[H_{2}O\right]=\frac{1\times 1000}{18}=55.56$
$\therefore K=\frac{{10}^{-7}\times {10}^{-7}}{(55.56{)}^2}\approx 3.25\times {10}^{-18}$