Question

One litre of water contains ${10}^{-7}$ moles of $H^{+}$ ions. Degree of ionisation of water (in percentage) is:

• A. $1.8\times {10}^{-7}$
• B. $1.8\times {10}^{-9}$
• C. $3.6\times {10}^{-7}$
• D. $3.6\times {10}^{-9}$

Answer 1

Answer: (A)

$1.8\times {10}^{-7}$

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

Let $x$ be the degree of ionisation.

We know that molarity of water $=55.5M$

$\therefore 55.5\times x={10}^{-7}$

$x=\frac{{10}^{-7}}{55.5}=0.018\times {10}^{-7}$

$\because$ We have $\alpha$ consider it to be in $\%$ :

$\therefore x$ % $=0.018\times {10}^{-7}\times 100$

$=1.18\times {10}^{-7}$