1. A solution of .050 M BaCl 2 and .025 M CaCl 2. K sp of BaSO 4 is 1.1 x 10 -10
ID: 866255 • Letter: 1
Question
1. A solution of .050 M BaCl2 and .025 M CaCl2. Ksp of BaSO4 is 1.1 x 10-10, and Ksp of CaSO4 is 2.4 x 10-5. a. What is the greatest concentration of SO42- that will precipitate on cation while leaving the other in solution? b. Which salt BaSO4 or CaSO4 will have the greatest molar solubility? c. Calculate the molar solubility of the insoluble salt you selected in part b, assuming that solid has been added to pure water. 2. As a result of mineral erosion and biological activity, phosphate ion is common in natural waters where it often precipitates as insoluble salts such as Ca3(PO4)2. If [Ca2+]init= [PO43-]init=1.0 x 10-8 M in a given river, will Ca3(PO4)2 precipitate? The Ksp of Ca3(PO4)2 is 1.2 x 10-29.
Explanation / Answer
1.
a.
Given, Ksp of BaSO4 = 1.1 x 10^-10
0.050 M BaCl2 => [Ba^2+] = 0.050 M
Ksp = [Ba^2+][SO42-]
1.1 x 10^-10 = 0.050 [SO42-]
[SO4^2-] = 2.2 x 10^-9 M
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Given, Ksp of CaSO4 = 2.4 x 10^-5
0.025 M CaCl2 => [Ca^2+] = 0.025 M
Ksp = [Ca^2+][SO4^2-]
2.4 x 10^-5 = 0.025 [SO42-]
[SO4^2-] = 9.6 x 10^-4 M
Thus, SO4^2- ion at concentrations greater than [SO4^2-] = 2.2 x 10^-9 M precipitates BaSO4 leaving the CaSO4 in solution.
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b.
CaSO4 has greater molar solubility.
BaSO4 <-----------> Ba^2+ + SO4^2-
KSp = [Ba2+][SO42-]
1.1 x 10^-10 = x^2
x = 1.05 x 10^-5 M = Molar solubility
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CaSO4 <-----------> Ca^2+ + SO4^2-
KSp = [Ca2+][SO42-]
2.4 x 10^-5 = x^2
x = 4.89 x 10^-3 M = Molar solubility
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c.
CaSO4 <-----------> Ca^2+ + SO4^2-
KSp = [Ca2+][SO42-]
2.4 x 10^-5 = x^2
x = 4.89 x 10^-3 M = Molar solubility
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2.
Given, [Ca2+]init = [PO43-]init =1.0 x 10-8 M
Ca3(PO4)2 <------------> 3Ca^2+ + 2PO4^3-
Qsp = [Ca^2+]^3 [PO4^3-]^2
Qsp = [1.0 x 10-8]^3 [1.0 x 10-8]^2
Qsp = 1 x 10^-40
Given, Ksp = 1.2 x 10^-29
As, Qsp < Ksp, Ca3(PO4)2 will not precipitate.
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