Question

The degree of dissociation of pure water at $25{}^{o}C$ is found to be $1.8\times {10}^{-9}$. Find $K_{w}and{K}_d$ at $25{}^{o}C$ ?

• A. $3.24\times {10}^{-18}$ ; $5.83\times {10}^{-20}$
• B. $1.00\times {10}^{-14}$ ; $1.80\times {10}^{-16}$
• C. $1.80\times {10}^{-16}$ ; $1.00\times {10}^{-14}$
• D. $1.00\times {10}^{-14}$ ; $1.00\times {10}^{-14}$

Answer 1

The correct option is B $1.00\times {10}^{-14}$ ; $1.80\times {10}^{-16}$

Since $\alpha =1.8\times {10}^{-9}$
and for water concentration , $C=\frac{1000}{18}=55.56M$
$\left[H^{+}\right]=\left[O{H}^{-}\right]=C\alpha =55.56\times 1.8\times {10}^{-9}=1\times {10}^{-7}M$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=(1\times {10}^{-7}{)}^2=1\times {10}^{-14}$
and $K_d=\frac{\left[H^{+}\right][OH{]}^{-}}{\left[H_{2}O\right]}{K}_d=\frac{K_w}{\left[H_{2}O\right]}{K}_d=\frac{{10}^{-14}}{55.56}=1.8\times {10}^{-16}$
Hence, $K_w$ and $K_d$ are $1.00\times {10}^{-14}$ $1.8\times {10}^{-16}$ respectively.