Question

The degree of dissociation of water at ${25}^{o}C$ is $1.8\times {10}^{-7}$% and density is $1.0g{cm}^{-3}$. The ionic constant for water is:

• A. $1.0\times {10}^{-14}$
• B. $2.0\times {10}^{-16}$
• C. $1.0\times {10}^{-16}$
• D. $1.0\times {10}^{-8}$

Answer 1

Answer: (B)

$2.0\times {10}^{-16}$

For the equilibrium:

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

given $\alpha =1.8\times {10}^{-7}\times 1/100=1.8\times {10}^{-9}$

Since $\alpha$<<<1, $\left[H_{2}O\right]$ is not much affected.

therefore concentrations at equilibrium are as:

$\left[H_{2}O\right]=\frac{1000\times 1}{18}\approx 55.5M$

$\left[H^{+}\right]=\alpha \times 55.5\approx 1\times {10}^{-7}$

$\left[O{H}^{-}\right]=\alpha \times 55.5\approx 1\times {10}^{-7}$

Ionic constant of water: $K=\frac{\left[O{H}^{-}\right]\left[H^{+}\right]}{\left[H_{2}O\right]}$

on substitution we get

$K=\frac{{10}^{-7}\times {10}^{-7}}{55.5}$

$K=\frac{{10}^{-7}\times {10}^{-7}}{55.5}$

$K\approx 2\times {10}^{-16}$

Therefore option B is correct.