Question

The percentage degree of dissociation of water at ${25}^{o}C$ is $1.9\times {10}^{-7}$ and density is $1.0gc{m}^{-3}.$ The ionic constant for water is:

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

Answer 1

The correct option is B $1.0\times {10}^{-14}$

The dissociation reaction of water is $\begin{array}{cc}{H}_{2}O& \rightleftharpoons H^{+}+O{H}^{-}\\ (1-\alpha )c& \alpha c\alpha c\end{array}$
$\text{Given}:\alpha =1.9\times {10}^{-9}$
Density of water $=\frac{1.0g}{c{m}^3}$
$\therefore c=\frac{1}{18}\times 1000=55.56$ $mol{L}^{-1}$
$\therefore \left[H^{+}\right]=c\alpha =55.56\times 1.9\times {10}^{-9}\left[H^{+}\right]=1.055\times {10}^{-7}=[OH{]}^{-}$

$\therefore K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$K_w=(1.055\times {10}^{-7}{)}^2=1.0\times {10}^{-14}$
$K_w={10}^{-14}$ =Ionic product of water