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Calculate the pH of 1.0 × 10-8 M solution of HCl?

HCl is 100% ionised in water.( H₂O)

? [H⁺] = 1.0 × 10^-8 M due to addition of HCl to water.
  H₂O ? H⁺ + OH^-  

From the Auto dissociation of water,
        [H⁺] [OH^-] = 1.00 × 10^-14

Hence [H⁺] = 1.00 × 10^-7 M ( [OH]^- = 10^-7)
?  Total [H⁺] = 1.00× 10^-7 + 1.0 × 10^-8 M
                      (From H₂O)   (From HCl)

                   = 10^-8 (10 + 1) M
                [H⁺] = 11 × 10^-8  = 1.1 x  10^-7
And pH = - log[H⁺] = -log (1.1 × 10^-7)

           pH = 6.96