CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.43

A solution is saturated with Ag2SO4. (a) Calculate [Ag+] in this saturated solution. (b) What mass of Na2SO4 must be added to 0.500 L of the solution to decrease [Ag+] to 4.0×10–3 M.




Answer:

(a)
Ag2SO4(s)
2 Ag+(aq)
+
SO42–(aq)
Ksp = [Ag+]e2[SO42–]e = 1.4×10–5
Initial
0
0
Change
+ 2x
+ x
Equilibrium
2x
x

1.4×10–5 = [2x]2[x] = 4x3

x = 0.015 M = silver sulfate concentration in a saturated solution.

[Ag+] = 2x = 2(0.015) = 0.030 M.

(b)

Ag2SO4(s)
2 Ag+(aq)
+
SO42–(aq)
Ksp = [Ag+]e2[SO42–]e = 1.4×10–5
Initial
?
0
Change
?
+ x
Equilibrium
4.0×10–3
x

(Note: even though the initial concentration of silver ion is not known, the equilibrium concentration is given in the problem so a solution can be obtained.)

1.4×10–5 = [4.0×10–3]2[x]

x = 0.88 M = the concentration of sulfate required to reduce the silver ion concentration to the desired level.

The number of moles required in 0.500 L = (0.88 mol/L)(0.500 L) = 0.44 mol

The molar mass of Na2SO4 = 2(23.0) + 32.1 + 4(16.0) = 142.1 g/mol.

The number of grams required = (0.44 mol)(142.1 g/mol) = 63 g.