a) The equilibrium reaction is given as Ca 3 (PO 4 ) 2 (s) <======> 3 Ca 2+ (aq)
ID: 699345 • Letter: A
Question
a) The equilibrium reaction is given as
Ca3(PO4)2 (s) <======> 3 Ca2+ (aq) + 2 PO43- (aq)
b) Let x be the solubility of Ca3(PO4)2. Then, the solubility of Ca2+ = 3x and that of PO42- = 2x (note the 1:3:2 nature of dissociation). The expression for the solubility constant, Ksp is given as
Ksp = [Ca2+]3[PO43-]2 = (3x)3*(2x)2 = 27x3*4x2 = 108x5 (square braces denote molar concentrations or solubility; by convention, the molar concentration of a solid species is taken as unity).
c) Set up the ICE chart for the solubility of Ca3(PO4)2 as below.
Ca3(PO4)2 (s) <=======> 3 Ca2+ (aq) + 2 PO43- (aq)
initial - 0 0
change - +3x +2x
equilibrium - 3x 2x
We have assumed the initial concentration of Ca3(PO4)2 (s) to be unity. Ca3(PO4)2 is a sparingly soluble salt and hence, its concentration doesn’t change over time or when equilibrium is reached.
It is given that [Ca2+] = 7.69*10-6 M; therefore,
3x = 7.69*10-6 M
====> x = (7.69*10-6 M)/3 = 2.563*10-6 M 2.56*10-6 M.
The solubility of PO43- is given as [PO43-] = 2x = 2*2.56*10-6 M = 5.12*10-6 M (ans).
d) Plug in the values of [Ca2+] and [PO43-] in the Ksp expression and obtain
Ksp = [Ca2+]3[PO43-]2 = (7.69*10-6)3*(5.12*10-6)2 = 1.192*10-26 1.19*10-26 (ans).
e) Note that NA3PO4 is an ionic salt and ionizes completely, i.e, we will have
Na3PO4 (aq) -------> 3 Na+ + PO43- (aq)
Due to the 1:1 nature of dissociation, we must have [PO43-] = 0.123 M.
Note that [PO43-] from Na3PO4 is much higher than the concentration of PO43- produced due to ionization of Ca3(PO4)2. Consequently, we can assume PO43- to be furnished by Ca3(PO4)2 and write
Ksp = [Ca2+]3[PO43-]2
=====> 1.19*10-26 = [Ca2+]3*(0.123)2
=====> [Ca2+]3 = 7.86569*10-25
=====> [Ca2+] = 9.2309*10-9
The solubility of Ca2+ in 0.123 M Na3PO4 is 9.2309*10-9 M and the solubility of Ca3(PO4)2 will be 1/3*solubility of Ca2+ (1:3 nature of dissociation) = 1/3*(9.2309*10-9 M) = 3.0770*10-9 M 3.08*10-9 M (ans).
Explanation / Answer
ACTUAL TEST SCS Fall 2017 AP Chemistry FRQ on Ksp and pH Titration Name: Show all work. 1. In a saturated solution of calcium phosphate, Cas(PO4)2, the equilibrium concentration of [Ca2]-7.69 x106 M. a) Write the equilibrium chemical reaction for the Ksp for calcium phosphate, b) write the Ksp expression in terins of "X" = solubility of Ca(PO4)2. b) Set up a short ICE Box and calculate the equilibrium molar concentration of phosphate, [P043-]. c) What is the solubility product constant for calcium phosphate? 2 i) Calculate the solubility of calcium phosphate in 0.123 M NaPO+
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