Question

1. What is the pH of pure water?

2. What is the change in pH when 10.0 mL of 0.010 M HCl is added to 500 mL of pure water.

3. What is the change in pH when 10.0 mL of 0.010 M NaOH is added to 500 mL of pure water.

Answer 1

1) in pure water [H+]= [OH-]= 1.0 x10^-7

    pH= -log [H+] = log(1.0 x10^-7) =7

2) on addition of 10ml 0.01M HCl to 500ml water

    HCl millimoles = 10 x0.01 = 0.1

[H+] in the solution = HCl millimoles /total volume(ml)

                               = 0.1 / (10 + 500)

                                = 1.96 x10^-4

pH= -log[H+] = -log(1.96 x10^-4)

pH= 3.70

pH decrease from 7 to 3.70

3) on addition of 10ml 0.01M NaOH to 500ml water

    NaOH millimoles = 10 x0.01 = 0.1

[OH-] in the solution =    NaOH millimoles /total volume(ml)

                               = 0.1 / (10 + 500)

                                = 1.96 x10^-4

pOH= -log[OH-] = -log(1.96 x10^-4)

pOH= 3.70

pH= 14-pOH

   pH = 10.30

pH increases from 7 to 10.30