Calculate the $[\mathrm{OH}]$ if pOH of solution is 4.94
Calculate the $[\mathrm{OH}]$ if pOH of solution is 4.94
- $2.356 \times 10^{-5} \mathrm{M}$
- $1.881 \times 10^{-5} \mathrm{M}$
- $\quad 1.417 \times 10^{-5} \mathrm{M}$
- $1.148 \times 10^{-5} \mathrm{M}$
Solution
$\begin{array}{ll} & \mathrm{pOH}=-\log _{10}\left[\mathrm{OH}^{-}\right] \\ \therefore \quad & \log _{10}\left[\mathrm{OH}^{-}\right]=-4.94 \\ \therefore \quad & {\left[\mathrm{OH}^{-}\right]=10^{(-4.94)}=1.148 \times 10^{-5} \mathrm{M}}\end{array}$
Asked in: MHT CET 2024 (04 May Shift 2)
Practice more Ionic Equilibria questions on Aicharya