Question

How many moles of BaF2(s) will dissolve in 150. mL of 0.60 M NaF? For BaF2, Ksp= 2.45 ×10–5

Answer 1

NaF is soluble in water. Hence, 0.60 M of NaF will give 0.60 M F^- in water.

BaF₂(s) Ba²⁺(aq) + 2F^-(aq)

Initial concentration (M)                                                       0                 0.60

Change in concentration (M)                                               X                  2X

Equilibrium concentration (M)                                              X            (0.60 + 2X)

Given, K_sp= 2.45 ×10^–5

Now,

K_sp = [Ba²⁺][F^-]²

or, 2.45 ×10^–5 = (X)(0.60 + 2X)²

or, 2.45 ×10^–5 = (X)(0.60)²              [(0.60 + 2X) 0.60 as 2X << 0.60]

or, X = 6.81 x 10^-5

Therefore, the molar solubility of BaF₂ in 150. mL of 0.60 M NaF = 6.81 x 10^-5 M

Volume of the solution = 150 mL = 0.150 L

Hence, the moles of BaF₂ will dissolve = molarity x volume of the solution in liter

                                                              = 6.81 x 10^-5 M x 0.150 L

                                                              = 1.02 x 10^-5 mol