Question

how many litres of water must be added to 1L of an aqueous solution of HCl having pH=1 to make an aqueous solution of pH=2. explain me with a clear steps

Answer 1

$$\begin{gathered} Volumeoftheoriginalsolution=1L \\ pH=1 \\ -\log \left[H^{+}\right]=1 \\ \left[H^{+}\right]=A.\log (-1) \\ ={10}^{-1} \\ pHofthenewsolution=2 \\ -\log \left[H^{+}\right]=2 \\ \left[H^{+}\right]=A.\log (-2) \\ ={10}^{-2} \\ Volumeofthenewsolution=? \\ Aspervolumetricprinciple \\ {V}_1{M}_1=V_2{M}_2 \\ 1\times {10}^{-1}=V_2\times {10}^{-2} \\ {V}_2=\frac{1\times {10}^{-1}}{{10}^{-2}}=10 \\ Thatmeansthevolumeofwateraddedtotheoriginalsolutionof1L=10-1=9L \end{gathered}$$