Ch 16B /17B Lewis theory, Ksp t Introduction to Solubility and the Solubility Pr
ID: 1038316 • Letter: C
Question
Ch 16B /17B Lewis theory, Ksp t Introduction to Solubility and the Solubility Product Constant 8 of 27> Constants| Periodic Table Part A Learning Goal A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration of Ba2+ ion in the solution was found to be 7.52x103 M.Calculate Ksp for BaF2 Express your answer numerically. To leam how to calculate the solubility from Ksp and vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: View Available Hint(s) CaF2(s) Ca2 (aq)+2F (aq) At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is K,-[Ca2+ ] [F-12 Asp 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. Submit Part B The value of Ksp for silver carbonate. Ag2CO3, is 8.10x 10-12. Calculate the solubility of Ag2CO3 in grams per liter. Express your answer numerically in grams per liter View Available Hint(s) solubility g/L SubmitExplanation / Answer
A)
At equilibrium:
BaF2 <----> Ba2+ + 2 F-
s 2s
[Ba2+] = s = 7.52*10^-3 M
Ksp = [Ba2+][F-]^2
Ksp = (s)*(2s)^2
Ksp = 4(s)^3
Ksp = 4(7.52*10^-3)^3
Ksp = 1.701*10^-6
Answer: 1.70*10^-6
B)
At equilibrium:
Ag2CO3 <----> 2 Ag+ + CO32-
2s s
Ksp = [Ag+]^2[CO32-]
8.1*10^-12=(2s)^2*(s)
8.1*10^-12= 4(s)^3
s = 1.265*10^-4 M
Molar mass of Ag2CO3,
MM = 2*MM(Ag) + 1*MM(C) + 3*MM(O)
= 2*107.9 + 1*12.01 + 3*16.0
= 275.81 g/mol
Molar mass of Ag2CO3= 275.81 g/mol
s = 1.265*10^-4 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.265*10^-4 mol/L * 275.81 g/mol
s = 3.489*10^-2 g/L
Answer: 3.49*10^-2 g/L