Question

1. What is the pH of a solution that contains 8.025 x 10-4 M NaOH?

2. Calculate the pH for a 9.21 x 10-7 M solution of Ba(OH)2.

3. Calculate the pOH for a 5.541 x 10-6 M solution of Ba(OH)2.

4. Calculate the pH for a solution with a hydronium concentration of 4.52 x 10-3 M.

Answer 1

1. NaOH is a strong electrolyte and it can be dissociable as follow, NaOH = Na⁺(aq) + OH^-(aq). So, from 1 mole of NaOH substance 1 mole of OH^- would be formed. So, [OH^-] = 8.025E-4 and we know that pOH = -log[OH^-] or, pOH = -log(8.025E-4) = 3.095 (considering f tends to 1 so, a = c, where f is activity coefficient, a is activity and c is the concentration). Now, pH+pOH = pKw = 14 and pOH = 3.095 so pH = 14-3.095 = 10.905. So, pH of the solution is 10.905.

2. Ba(OH)₂ is a strong electrolyte and it can be dissociable as follow, Ba(OH)₂ = Ba²⁺(aq) + 2 OH^-(aq) so from one mole of barium hydroxide 2 moles of OH^- would be formed. Now concentration of Ba(OH)₂ is 9.21E-7 so [OH^-] = (2×9.21E-7) = 1.842E-6 so, pOH = -log(1.842E-6) = 5.73 so, pH = 14-5.73 = 8.27. So, the pH of the solution is 8.27.

3.  Ba(OH)₂ is a strong electrolyte and it can be dissociable as follow, Ba(OH)₂ = Ba²⁺(aq) + 2 OH^-(aq), so from one mole of barium hydroxide 2 moles of OH^- would be formed. Now concentration of Ba(OH)₂ is 5.541E-6 so [OH^-] = (2×5.541E-6) = 1.108E-5, so pOH = -log(1.108E-5) = 4.95. So, pOH of the solution is 4.95.

4. The molecular formula of hydronium ion is H3O+ Now, we can assume this hydronium ion as H₂O + H⁺ so, the concentration of hydronium ion should be similar to the concentration of H⁺ ion of the solution. Now, [H₃O⁺] = 4.52E-3 so the pH = -log(4.52E-3) = 2.34 or the pH of the solution is 2.34.