Question

If a solution has twice as many hydroxide ions as in pure water at $25º\mathrm{C}$, then the ph of the solution at the same temperature will be

Answer 1

Step 1: Finding the concentration of ${\mathrm{OH}}^{-}$ions

Water dissociates into Hydrogen ions and Hydroxide ions:

$\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\rightleftharpoons \underset{\mathrm{Hydrogen}\mathrm{ions}}{{\mathrm{H}}^{+}}+\underset{\mathrm{Hydroxide}\mathrm{ions}}{{\mathrm{OH}}^{-}}$

$\therefore {\mathrm{K}}_{\mathrm{w}}=1\times {10}^{-14}\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{H}}^{+}\right]$

Since water is neutral so,

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\left[{\mathrm{H}}^{+}\right] \\ \Rightarrow 1\times {10}^{-14}=\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{H}}^{+}\right] \\ \Rightarrow \left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-7}\mathrm{mol}{\mathrm{L}}^{-1} \end{gathered}$$

Step 2: Finding the pH

Now according to the question, the solution has twice as many hydroxide ions in pure water at $25º\mathrm{C}$,

$$\begin{gathered} \therefore \left[{{\mathrm{OH}}^{-}}_{\mathrm{solution}}\right]=2\times \left[{{\mathrm{OH}}^{-}}_{\mathrm{water}}\right]=2\times {10}^{-7}\mathrm{mol}{\mathrm{L}}^{-1} \\ \Rightarrow \mathrm{pOH}=-\log \left[{{\mathrm{OH}}^{-}}_{\mathrm{solution}}\right] \\ \Rightarrow \mathrm{pOH}=-{\log }_{10}(2\times {10}^{-7}) \\ \Rightarrow \mathrm{pOH}=-{\log }_{10}2+7{\log }_{10}10 \\ \Rightarrow \mathrm{pOH}=-.301+7[\because \log 10=1] \\ \Rightarrow \mathrm{pOH}=6.69 \end{gathered}$$

Now pH of the solution will be:

$$\begin{gathered} \mathrm{pOH}+\mathrm{pH}=14 \\ \Rightarrow \mathrm{pH}=14-\mathrm{pOH} \\ \Rightarrow \mathrm{pH}=14-6.69 \\ \Rightarrow \mathrm{pH}=7.31 \end{gathered}$$

Therefore, the pH of the solution is $7.31$