Question

The dissociation of water at ${25}^{\circ }$C is $1.9\times {10}^{-7}\%$ and the density of water is $1.0$ g/cc. The ionisation constant of water is :

• A. $3.42\times {10}^{-6}$
• B. $3.42\times {10}^{-8}$
• C. $1.00\times {10}^{-14}$
• D. $2.00\times {10}^{-16}$

Answer 1

Answer: (D)

The degree of dissociation $\left(\alpha \right)$ for water is $\frac{1.9\times {10}^{-7}}{100}=1.9\times {10}^{-9}$.

$1$ L of water (molecular mass $=18$ g/mol) weighs $1000$ g.
Hence, concentration of water $=\frac{1000}{18}$ M.

The expression for the ionization constant of water is $K=\left[H^{+}\right]\left[O{H}^{-}\right]=C(\alpha {)}^2$.

Substituting values in the above expression, we get
$K=\frac{1000}{18}\times (1.9\times {10}^{-9}{)}^2=2.0\times {10}^{-16}$.