Question

Calculate the pH of the following aqueous solutions at 25°C. K_w at 25°C is 1.01e-14.

(a)    2.9 ✕ 10⁻¹¹M KOH (HINT: Does it make sense to get an acidic pH for this? A neutral solution may have a higher concentration of OH- than this solution.)
1

(b)    9.1 ✕ 10⁻⁷M HNO₃
2

Answer 1

1. Since the OH^- concentration (KOH ) is too low, the contribtion from dissoaciation of water must be consider while calculating pH.

             Total [OH^-] = (2.9 x 10^-11 + 10^-7 )M = 1 x 10^-7 M

             Given that, Ionic product of water, K_W = [H⁺][OH^-] = 1 x 10^-14

             Or, [H⁺] = (1 x 10^-14 ) – (1 x 10^-7) = 1 x 10^-7 M

              pH = -log [H⁺] = - log (1 x 10^-7 ) = 7 (i.e.- neutral)

b) 9.1 ✕ 10⁻⁷M HNO₃ = [H⁺]

             Total [H⁺] = (1 x 10^-7 ) + (9.1 ✕ 10⁻⁷) = 10.1 x 10^-7 M

             pH = -log [H⁺] = -log (10.1 x 10^-7) = 5.99