Question

Calculate the pH of a solution with [H₃O+] = 1.0 x 10^-8 M

please explain

Answer 1

Given [H₃O⁺] = 1.010 ^-8 M

We know that pH = -log [H₃O⁺]

                         = -log(1.010 ^-8 )

                        = -[log1.0+ log 10 ^-8 )                 Since Log AB = log A + Log B

                       = - log 1.0 -(-8log 10 )                 Since log A^x = x log A

                       = 0 + (8x1)

                      = 8      

which is abuse since pH of an acidic solution should not exceeds 7

Sine the [H₃O⁺] concentration is very weak

Total But the [H₃O⁺]_total = [H₃O⁺] _{from the acid} + [H₃O⁺] _{from water}

                               = 1.010 ^-8 M + 1.010 ^-7 M

                                 = 1.010 ^-7 (0.1+1)

                          = 1.1 10 ^-7 M

So pH = -log [H₃O⁺]_total

         = -log(1.110 ^-7 )

         = -[log1.1+ log 10 ^-7 )

         = - log 1.1 +7 log 10

         = -0.04 + 7

         = 6.96