Question #92436
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.
Expert's answer
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.
