Question

If the degree of dissociation of water at ${70}^{o}C$ is $1.28\times {10}^{-8}$. Then what is the ionic product of water at this temperature?

• A. $25\times {10}^{-16}{M}^2$
• B. $1.52\times {10}^{-14}{M}^2$
• C. $1.28\times {10}^{-8}{M}^2$
• D. $5.1\times {10}^{-13}{M}^2$

Answer 1

The correct option is D $5.1\times {10}^{-13}{M}^2$

Degree of dissociation of water $\alpha =1.28\times {10}^{-8}$
and concentration of water is$=\frac{1000}{18}=55.55M$

Equilibrium concentrations:
$H_{2}O\left(l\right)\rightleftharpoons H^{+}(aq.)+O{H}^{-}(aq.)C(1-\alpha )C\alpha C\alpha$

Dissociation constant,
$K_c=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}{K}_w=K_c\times \left[H_{2}O\right]=C\alpha \times C\alpha K_w=C^2{\alpha }^2{K}_w=(55.55{)}^2\times (1.28\times {10}^{-8}{)}^2=5.1\times {10}^{-13}$
Ionic product, $K_w=5.1\times {10}^{-13}{M}^2$