Question

Calculate the concentration of ${OH}^{-}$ ions in solution when $\left[H^{+}\right]=6.2\times {10}^{-2}mol{L}^{-1}$.

Answer 1

We know, $\left[H^{+}\right]\times \left[O{H}^{-}\right]=1\times {10}^{-14}$ at ${25}^{o}C$
(Given:$\left[H^{+}\right]=6.2\times {10}^{-2}mol{L}^{-1}$)
Thus,
$\left[O{H}^{-}\right]=\frac{1\times {10}^{-14}}{6.2\times {10}^{-2}}=1.61\times {10}^{-13}mol{L}^{-1}$
$H_{2}O⟶{OH}^{-}+H^{+}$
$\left[O{H}^{-}\right]=\frac{\left[H_{2}O\right]}{\left[H^{+}\right]}$