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 mBa(OH)2MBa(OH)2 

⇒ nHO- = 2 x 0.0225171.34 = 2.63 x 104 mol


[HO-] = nHO-V = 1.05 x 10-3 M


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

⇒ [H3O+] = 1.0 × 10-14HO- = 9.52 x 10-12 M

⇒ pH = 11.0