Exercise 2 | The Properties of Acids and Bases

What is [H3O+] in a 0.050 M solution of Ca(OH)2 at 25°C?

At 25°C, Kw = [H3O+][HO-] = 1.0 x 10-14 2


Ca(OH)(aq) → Ca2+ (aq) + 2 HO(aq)

From the stoichiometry of the reaction: nHO- = 2 nCa(OH)2

⇒ [HO-] = 2 [Ca(OH)2] = 0.10 M


[H3O+][HO-] = 1.0 x 10-14

⇒ [H3O+] = 1.0 × 10-141.0 × 10-1 = 1.0 x 10-13 M