Question

One liter of water ( molecular weight $ 18.06 $ ) weighs $ 0.9970 \,kg $ . The degree of ionisation of water is............if $ K_w = 1.10 \times 10^{-14} $ at $ 25^{\circ}C $

• A. $ 1.05 \times 10^{-7} $
• B. $ 1.9 \times 10^{-9} $
• C. $ 1.01 \times 10^{-11} $
• D. $ 4.52 \times 10^{-7} $

Answer 1

Answer: (B)

Given $K_w = 1.10 \times 10^{- 14}$
The value of $K_w$ is temperature dependent as it is an equilibrium constant.
The molarity of pure $H_2O$
$ = \frac{\text{Density}}{M}$
$ = \left(\frac{1000\,gL^{-1}}{18.0\,g\,mol^{-1}}\right)$
$ = 55.55\,M$
$\alpha = \frac{10^{-7}}{C}$
$ = \frac{10^{-7}}{55.6}$
$ = 1.9 \times 10^{-9}$