Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given, aqueous solution of$\mathrm{\text{N}}{\mathrm{\text{H}}}_3\left(\mathrm{\text{weak base}}\right)\text{ C}=0.01$M

$\mathrm{\text{α}}=5\text{\%}=\frac{5}{100}$

$\left[\mathrm{\text{O}}{\mathrm{\text{H}}}^{-}\right]=\text{C}\mathrm{\text{α}}=0.01\times \frac{5}{100}=5\times {10}^{-4}$

$\mathrm{\text{p}}\mathrm{\text{O}}\mathrm{\text{H}}=-\mathrm{\text{log}}\left[\mathrm{\text{O}}{\mathrm{\text{H}}}^{-}\right]$

$=\mathrm{\text{log}}\left(5\times {10}^{-4}\right)$

$=4\mathrm{\text{log}}10-\mathrm{\text{log}}5$

$=4-0.6989$

$=3.3010$

$\therefore$$\mathrm{\text{p}}\mathrm{\text{H}}=14-\mathrm{\text{p}}\mathrm{\text{O}}\mathrm{\text{H}}=14-3.3010=10.6990$