Exercise 3 | Acids and Bases

Ba(OH)₂ (aq) → Ba²⁺ (aq) + 2 HO^- (aq)

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

⇒ n_HO- = 2 x $\frac{{\mathrm{m}}_{\mathrm{Ba}\left(\mathrm{OH}\right)2}}{{\mathrm{M}}_{\mathrm{Ba}\left(\mathrm{OH}\right)2}}$ 

⇒ n_HO- = 2 x $\frac{0.0225}{171.34}$ = 2.63 x 10⁴ mol

[HO^-] = $\frac{{\mathrm{n}}_{{\mathrm{HO}}^{-}}}{\mathrm{V}}$ = 1.05 x 10^-3 M

At 25°C, K_w = [H₃O⁺][HO^-] = 1.00 x 10^-14 M ²

⇒ [H₃O⁺] = $\frac{1.0 \times {10}^{-14}}{\left[{\mathrm{HO}}^{-}\right]}$ = 9.52 x 10^-12 M

⇒ pH = 11.0