Question

Given the reactions: YFz(s) = Y3+(aq) + 3F (aq) Ksp = 8.62 x 10-21 HF(s) = H*(aq) + F(aq) Kg = 6.3 x 10-4 a) How does the solubility of YF, change with increasing pH of the solution? b) How does the solubility of YF, change with decreasing pH of the solution?

Answer 1

a) The solubility of YF3 would decrease with increasing pH of the solution. This is because the fluoride ion concentration would decrease with increasing pH, according to the equilibrium: F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq). The lower the concentration of fluoride ions in the solution, the less YF3 can dissolve, based on the solubility equilibrium.

b) The solubility of YF3 would increase with decreasing pH of the solution. This is because the fluoride ion concentration would increase with decreasing pH, according to the equilibrium: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq). The higher the concentration of fluoride ions in the solution, the more YF3 can dissolve, based on the solubility equilibrium.