Question

What is dissociation constant of water at ${25}^{o}C$ from the following data:
Conductivity of $H_{2}O=5.8\times {10}^{-8}Sc{m}^{-1}$, ${\lambda }_{H^{+}}^{\infty }=350.0$ and ${\lambda }_{O{H}^{-}}^{\infty }=198.0Sc{m}^{2}mo{l}^{-1}$

• A. $3.8\times {10}^{-16}mol/litre$
• B. $2\times {10}^{-16}mol/litre$
• C. $1.8\times {10}^{-16}mol/litre$
• D. None of these

Answer 1

The correct option is C $1.8\times {10}^{-16}mol/litre$

Suppose water contains X mol/litre or X eq/litre of $H^{+}$ ions or $O{H}^{-}$ ions.
$\because$ Equivalent of $H^{+}=$ Eq. of $H_2$
$\therefore$ Volume of $H_{2}O$ in mL having 1 Eq. $H^{+}=\frac{1000}{X}$
$\therefore$ Equivalent conductivity of water, $A_{eq.}^{\infty }{H}_{2}O=\frac{k\times 1000}{N}$
$=5.8\times {10}^{-8}\times \frac{1000}{X}$
Since, water dissociated feebly, i.e., water may be considered to be a dilute solution of $H^{+}$ and $O{H}^{-}$ ions.
$\therefore A_{eq.H_{2}O}^{\infty }={\lambda }_{H^{+}}^{\infty }+{\lambda }_{O{H}^{-}}^{\infty }$
$\therefore 5.8\times {10}^{-8}\times \frac{1000}{X}=350+198=548$
$\therefore X=1.0\times {10}^{-7}$
$\therefore \left[H^{+}\right]=\left[O{H}^{-}\right]=1\times {10}^{-7}$
For the equilibrium
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
Equilibrium constant $\left(K\right)=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}$
$K_w=K\times \left[H_{2}O\right]=\left[H^{+}\right]\left[O{H}^{-}\right]$
$=1.0\times {10}^{-7}\times 1.0\times {10}^{-7}=1\times {10}^{-14}$
$\therefore K=\frac{K_w}{\left[H_{2}O\right]}=\frac{1\times {10}^{-14}}{55.5}$ $\left[\right[H_{2}O]=\frac{1000}{18}=55.5mol/litre]$
$=1.8\times {10}^{-16}mol/litre$