Question

Calculate the pH of a solution formed by mixing 50.00 mL of 0.0100 M HNO3 with 30.00 mL of 0.0150 M solution of Ba(OH)2

Answer 1

50.00 mL of 0.0100 M of HNO₃ = 50.00 mL x 0.0100 M = 0.500 mmol of HNO₃

HNO₃ is a strong acid. 0.500 mmol of HNO₃ gives 0.500 mmol of H₃O⁺ in the aqueous solution.

30.00 mL of 0.0150 M of Ba(OH)₂ = 30.00 mL x 0.0150 M = 0.450 mmol of Ba(OH)₂

Ba(OH)₂ is a strong base. 0.450 mmol of Ba(OH)₂ gives 2 x 0.450 mmol = 0.900 mmol of OH^- in the aqueous solution.

Now, on mixing, 0.500 mmol of H₃O⁺ neutralizes 0.500 mmol of OH^-.

Hence, the excess OH^- in the solution = (0.900 - 0.500) mmol = 0.400 mmol

Total volume of solution after mixing = (50.00 + 30.00) = 80.00 mL

Thus, the concentration of OH^- in the solution = 0.400 mmol/80.00 mL

                                                                         = 0.00500 M

Now,

pH = 14 - pOH

      = 14 + log[OH^-]

      = 14 + log(0.00500)

      = 14 - 2.3

      = 11.7

Therefore, the pH of the solution = 11.7