Question

At ${25}^{\circ }\text{C}$, the degree of ionization of water was found to be $1.8\times {10}^{-9}$. Calculate the ionic product of water at this temperature.

• A. $1\times {10}^{-14}{\text{M}}^2$
• B. $1\times {10}^{-7}{\text{M}}^2$
• C. $5\times {10}^{-14}{\text{M}}^2$
• D. $5\times {10}^{-7}{\text{M}}^2$

Answer 1

The correct option is A $1\times {10}^{-14}{\text{M}}^2$

If $\alpha$ is the degree of dissociation of water, then we have
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
$\left[H^{+}\right]=\left[O{H}^{-}\right]=c\alpha$
If mass of $1{\text{dm}}^3$ water is taken as $1000\text{g}$, then
$c=\frac{n}{V}=\frac{m/M}{V}=\frac{\left(1000\text{g}\right)/\left(18{\text{g mol}}^{-1}\right)}{1{\text{dm}}^3}=55.56\text{M}$
Thus, $K_i=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}=\frac{(c\alpha {)}^2}{c(1-\alpha )}\approx c{\alpha }^2(\text{assuming}\alpha <<1)$
$\Rightarrow \left(55.56\text{M}\right)(1.8\times {10}^{-9}{)}^2=1.8\times {10}^{-16}\text{M}$
and $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=(c\alpha {)}^2=[\left(55.56\text{M}\right)(1.8\times {10}^{-9}){]}^2$
$=1.0\times {10}^{-14}{\text{M}}^2$