Exercise 3 | The Properties of Acids and Bases

What is the pH of 250 mL of a solution containing 0.0225 g of Ba(OH)2 at 25°C?

Ba(OH)(aq) → Ba2+ (aq) + 2 HO(aq)

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

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

⇒ nHO- = 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, Kw = [H3O+][HO-] = 1.00 x 10-14 2

⇒ [H3O+] =  = 9.52 x 10-12 M

⇒ pH = 11.0