Question

A 25.0 mL aliquot of a saturated solution of magnesium hydroxide (Mg(OH)2) is titrated with 0.00062 M HCl. The student began his titration when the buret read 1.36 mL and the reaction reached the equivalence point when the buret read 12.65 mL of HCl.

a. How many moles of hydroxide ions were neutralized?

b. What was the molar concentration of the hydroxide ions in the saturated solution?

c. What is the molar solubility of magnesium hydroxide?

d. Calculate Ksp for Mg(OH)2.

Answer 1

volume of HCl used for titration = (12.65 - 1.36) = 11.29 ml = 0.01129 L

mole of HCl = 0.01129 L * 0.00062 mole / L = 7.0 * 10^-6 mole.

Mg(OH)2 + 2 HCl ..............> MgCl2 + 2 H2O

a) moles of hydroxide ions = 7.0 * 10^-6 mole.

b)  molar concentration of the hydroxide ions in the saturated solution = 7.0 * 10^-6 mole / 0.025 L = 2.80 * 10^-4 M

c) [Mg^2+] = 2.80 * 10^-4 M / 2 = 1.40 * 10^-4 M

molar solubility of Mg(OH)2 = 1.40 * 10^-4 M

d) Ksp = [Mg^2+][OH-]^2 = (1.40 * 10^-4) * (2.80 * 10^-4)^2 = 1.10 * 10^-11