##### 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.

(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.