The pH of $0.005 \mathrm{M} \mathrm{KOH}$ is 9.95. Calculate the $\left[\mathrm{OH}^{-}\right]$

The pH of $0.005 \mathrm{M} \mathrm{KOH}$ is 9.95. Calculate the $\left[\mathrm{OH}^{-}\right]$
  1. $6.71 \times 10^{-4} \mathrm{M}$
  2. $1.12 \times 10^{-4} \mathrm{M}$
  3. $4.45 \times 10^{-5} \mathrm{M}$
  4. $8.91 \times 10^{-5} \mathrm{M}$

Solution

$\begin{aligned} & \mathrm{P}^{\mathrm{H}}+\mathrm{P}^{\mathrm{OH}}=14 \\ & \mathrm{P}^{\mathrm{OH}}=14-.9 .95 \\ & =4.05 \\ & \mathrm{P}^{\mathrm{OH}}=-\log _{10}\left[\mathrm{OH}^{-}\right] \\ & {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{P}^{\mathrm{OH}}}=10^{-4.05}} \\ & =10^{-5+0.95}=10^{-5} \times 10^{0.95}=8.91 \times 10^{-5} \mathrm{M}\end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

Practice more Ionic Equilibria questions on Aicharya