pH=−log[H+] H+=10−8mol/L
So, in aqueous solution, in H+ion concentration of acid, H+ion concentration of water is also added.
Hence, total [H+]of acid =(10−8+10−7)mol/L =10−8(1+10) =11×10−8mol/L
Therefore, pH=−log[11×10−8] =−log11+8log10 =−1.0414+8 pH=6.9586