Question

the pH of solution at 25 degree celsius which has twice as many hydroxide ion as in pure water at 25 degree celsius will be ??

Answer 1

$$\begin{gathered} \text{Dear Student,} \\ \left[{\mathrm{OH}}^{-}\right]=2\left[{\mathrm{OH}}^{-}\right]\mathrm{at}{25}^{°}\mathrm{C}\mathrm{of}\mathrm{pure}\mathrm{water} \\ \mathrm{At}{25}^{°}\mathrm{C},\left[{\mathrm{OH}}^{-}\right]\mathrm{of}\mathrm{pure}\mathrm{water}={10}^{-7}\mathrm{M} \\ \left[{\mathrm{OH}}^{-}\right]=2\times {10}^{-7}\mathrm{M} \\ \mathrm{Kw}=\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{H}}^{+}\right] \\ {10}^{-14}=2\times {10}^{-7}\mathrm{M}\left[{\mathrm{H}}^{+}\right] \\ \left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-14}}{2\times {10}^{-7}}=\frac{{10}^{-7}\times 10}{2\times 10}=5\times {10}^{-8} \\ \left[{\mathrm{H}}^{+}\right]=5\times {10}^{-8} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log 5\times {10}^{-8} \\ \mathrm{pH}=5.591 \\ \mathrm{Regards} \end{gathered}$$