***CHEMISTRY AQUEOUS EQUILIBRIA QUESTION***PLEASE HELP*** PLEASE ANSWER CAREFULL
ID: 894655 • Letter: #
Question
***CHEMISTRY AQUEOUS EQUILIBRIA QUESTION***PLEASE HELP***
PLEASE ANSWER CAREFULLY AND MAKE SURE TO DO ALL PARTS!
± Introduction to Solubility and the Solubility Product Constant Learning Goal: To learn 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: 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 Ksp 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.Explanation / Answer
BaF2(s) <----> BaF2(aq) <----> Ba2+(aq)+ 2 F-(aq)
Ksp = [Ba2+][F-]^2
[Ba2+] = 7.52*10^(3-) M
[F-] = 2*7.52*10^(3-) = 0.01504
Ksp = (7.52*10^(-3)*(0.01504^2) = 1.701*10^-6 M^3
Part B
K2CO3(s) ----> 2K+(aq) + CO3^2-(aq)
Ksp = (2s)^2(s) = 4s^3
s = solubility = (Ksp/4)^(1/3)
= ((8.10*10^(-12)/4)^(1/3)
= 0.000126
= 1.26*10^(-4) g/L
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