Question

Percentage ionization of $H_{2}O$ at certain temperature is $3.6\times {10}^{-7}\%$. Calculate $K_w$ at this temperature.

• A. $K_w=4\times {10}^{-14}$
• B. $K_w=3\times {10}^{-14}$
• C. $K_w=1\times {10}^{-14}$
• D. none of these

Answer 1

The correct option is D none of these

Actual ionisation of water at certain temperature$=\frac{\%ofionization}{100}=\frac{3.6\times {10}^{-7}}{100}=3.6\times {10}^{-9}$

$2H_{2}O⟶O{H}^{-}+H_3{O}^{+}$

initial $1$ $0$ $0$

Eqb. $(1-3.6\times {10}^{-9})$ $(3.6\times {10}^{-9})$ $(3.6\times {10}^{-9})$

$K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]$

$=(3.2\times {10}^{-9}{)}^2$

$K_w=12.96\times {10}^{-18}$