Question

If the degree of dissociation of water at ${90}^{o}C$ is $1.28\times {10}^{-8}$ then the ionization constant of water at ${90}^{o}C$ is:

• A. $7.52\times {10}^{-12}M$
• B. $9.07\times {10}^{-15}M$
• C. $1.28\times {10}^{-14}M$
• D. $1.38\times {10}^{-14}M$

Answer 1

Answer: (A)

$7.52\times {10}^{-12}M$

Degree of dissociation of water $\alpha =1.28\times {10}^{-8}$

$H_{2}O\rightleftharpoons H^{+}+{OH}^{-}$

$C(1-\alpha )$ $C\alpha$ $C\alpha$

Dissociation constant, $K_w=\frac{\left[H^{+}\right]\left[{OH}^{-}\right]}{\left[H_{2}O\right]}=C\alpha .C\alpha \left\{[H_{2}O\approx 1]\right\}$

$=C^2{\alpha }^2$

$\approx {\alpha }^2[C^2\approx 1]$