Exercise 2 | Acids and Bases

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

Ca(OH)₂ (aq) → Ca²⁺ (aq) + 2 HO^- (aq)

From the stoichiometry of the reaction: n_HO- = 2 n_Ca(OH)2

⇒ [HO^-] = 2 [Ca(OH)₂] = 0.10 M

[H₃O⁺][HO^-] = 1.0 x 10^-14

⇒ [H₃O⁺] = $\frac{1.0 \times {10}^{-14}}{1.0 \times {10}^{-1}}$ = 1.0 x 10^-13 M