To learn how to calculate the solubility from K_sp and vice versa. Consider the

Question

To learn how to calculate the solubility from K_sp and vice versa.
Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution:

m CaF_2{(s)} ightleftharpoons m Ca^{2+}{(aq)}+2 F^-(aq)
At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid m CaF_2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is

K_{ m sp}=[ m Ca^{2+}][ m F^-]^2
K_sp is called the solubility product and can be determined experimentally by measuring the solubility, which is the amount of compound that dissolves per unit volume of saturated solution.
A saturated solution of barium fluoride, m BaF_2 , was prepared by dissolving solid m BaF_2 in water. The concentration of m Ba^{2+} ion in the solution was found to be 7.52

Explanation / Answer

helpfull The concentration of Pb2+ ion = 2.08*10^-3 Write the equation for the dissociation of PbF2 PbF2 ? Pb 2+ + 2F- Ksp = [Pb2+] * [ Cl-]² If the concentraion of Pb2+ is 2.08*10^-3M, as given in the question, then [F-]= 2.08*10^-3 *2 = 4.16*10^-3 Substituting: Ksp = [2.08*10^-3]*[4.16*10^-3]² Ksp = 3.5996*10^-8 Ksp = 3.60*10^-8

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