Question

Calculate the dissociation constant of water at ${25}^{\circ }C$ from the following data.

Specific conductance of $H_{2}O=5.8\times {10}^{-8}mhoc{m}^{-1},:{\lambda }_{H^{+}}^{\infty }=350.0$ and ${\lambda }_{O{H}^{-}}^{\infty }=198.0mhoc{m}^2$

The ans is $k\times {10}^{-16}mole/litre$. find k

• A. $1.8$
• B. $100$
• C. $1$
• D. $0.01$

Answer 1

The correct option is B $100$

${\Lambda }_{H_{2}O}^{\infty }={\lambda }_{H^{+}}^{\infty }+{\lambda }_{O{H}^{-}}^{\infty }=350.0+198.0=548.0mhoc{m}^2$
$\Lambda =\frac{1000\kappa }{C}$
$548=\frac{1000\times 5.8\times {10}^{-8}}{C}$
$C=1.0\times {10}^{-7}$
The dissociation constant of water is $C^2=(1.0\times {10}^{-7}{)}^2=1.0\times {10}^{-14}=k\times {10}^{-16}$
Hence, $k=100$