Question

If the ionic product of water at ${40}^{o}C$ is $4\times {10}^{-14}$. The concentration of $O{H}^{-}$ ions in water at ${40}^{\circ }C$ is:

• A. $2\times {10}^{-9}M$
• B. $4\times {10}^{-7}M$
• C. $4\times {10}^{-3}M$
• D. $2\times {10}^{-7}M$

Answer 1

The correct option is D $2\times {10}^{-7}M$

$H_{2}O\left(l\right)\rightleftharpoons H^{+}(aq.)+O{H}^{-}(aq.)K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$4\times {10}^{-14}=\left[H^{+}\right]\times \left[O{H}^{-}\right]$
At equilibrium,
$\left[H^{+}\right]=\left[O{H}^{-}\right]\therefore K_w=\left[O{H}^{-}{]}^2\right[O{H}^{-}]=\sqrt{K_w}=\sqrt{4\times {10}^{-14}}[O{H}^{-}]=2\times {10}^{-7}M$