Question 16.12: Calculate the solubility of MgF2 in water at 25 °C in units ......
Calculate the solubility of MgF_{2} in water at 25 °C in units of:
(a) Moles per liter (b) Grams per liter
STRATEGY
(a) Write the balanced equation for the solubility equilibrium assuming the complete dissociation of MgF_{2} and set up an equilibrium table. If we define x as the number of moles per liter of MgF_{2} that dissolves, then the saturated solution contains x mol/L of Mg^{2+} and 2x mol/L of F^{-}. Substituting these equilibrium concentrations into the expression for K_{sp} and solving for x gives the molar solubility.
(b) To convert the solubility from units of moles per liter to units of grams per liter, multiply the molar solubility of MgF_{2} by its molar mass (62.3 g/mol).
| IDENTIFY | |
| Known | Unknown |
| K_{sp} at 25 °C (7.4\times 10^{-11}) (Appendix C) | Solubility (M, g/L) |
(a) The equilibrium table for the dissolution of MgF_{2} is:
Table 1
Substituting the equilibrium concentrations into the expression for K_{sp} gives
K_{sp}=7.4\times 10^{-11}=[Mg^{2+}][F^{-}]^{2}=(x)(2x)^{2}4x^{3}=7.4\times 10^{-11}
x^{3}=1.8\times 10^{-11}
x=[Mg^{2+}]= Molar solubility =2.6\times 10^{-4} mol/L.
Thus, the molar solubility of MgF_{2} in water at 25 °C is 2.6\times 10^{-4} M.
Note that the number 2 appears twice in the expression (x)(2x)². The exponent 2 is required because of the equilibrium equation, K_{sp}= [Mg^{2+}] [F^{-}]^{2}. The coefficient 2 in 2x is required because each mole of MgF_{2} that dissolves gives 2 mol of F^{-}(aq).
(b) Solubility (in g/L) =\frac{2.6\times 10^{-4} mol}{L}\times\frac{62.3 g}{mol}=1.6\times 10^{-2} g/L
Table 1
| Solubility Equilibrium | MgF_{2}(s) | \xrightleftharpoons{} | Mg^{2+}(aq) | + | 2 F^{-}(aq) |
| Initial concentration (M) | 0 | 0 | |||
| Change (M) | + x | + 2x | |||
| Equilibrium concentration (M) | x | 2x |