Answer to Question #92436 in Chemistry for Tj

Question

Given the solubility of MgF2(s) is 8.4 x 10-7 M, calculate the value of Ksp for the reaction, MgF2(s)Mg2+(aq) + 2 F-(aq) at 275 oC.

Answer 1

8.4×10^-7 M   X M            Y M

  MgF₂(s) = Mg²⁺(aq) + 2F^-(aq)

   1 M           1 M           2M

[Mg²⁺(aq)] = X = 8.4×10^-7 M.

[F^-(aq)] = Y = 16.8×10^-7 M.

Ksp =  [Mg²⁺(aq)] × [F^-(aq)]² = 8.4×10^-7 × (16.8×10^-7)² = 2.37×10^-18.