Calculate the pH of the solution.

Question

A chemist dissolves 159. mg of pure barium hydroxide in enough water to make up 80. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 

25 C.)

Round your answer to 2 significant decimal places. 

Answer 1

Barium hydroxide = Ba(OH)₂

Ba(OH)₂ ==> Ba²⁺ + 2OH^- (note there are 2 OH^- for each Ba(OH)₂)

molar mass Ba(OH)₂ = 171.3 g / mole

moles Ba(OH)₂ = 159. mg x 1 g / 1000 mg x 1 mol / 171.3 g = 9.282x10-4 moles Ba(OH)₂

moles OH- = 9.282x10^-4 mols Ba(OH)₂ x 2 mols OH- / mol Ba(OH)₂ = 1.856x10-3 moles OH-

molarity of OH- = 1.856x10^-3 moles / 80. ml x 1000 mls / L = 0.0232 mols / L = 0.0232 M

Next, find pOH, and then pH:

pOH = -log [OH-] = -log 0.0232 = 1.63

pH = 14 - pOH

pH = 14 - 1.63

pH = 12.37 = 12 (2 sig.figs. based on 80. ml having only 2 s.f.)