A 100.0 mL solution is 0.0100M in Ba(NO 3 ) 2 and 0.0100M in Pb(NO 3 ) 2 . To th

Question

A 100.0 mL solution is 0.0100M in Ba(NO3)2 and 0.0100M in Pb(NO3)2. To this solution was added a slight excess of SO42- over the Ba2+ consisting of 101.0 mL of 0.0100M H2SO4. Under the conditions present,

Ksp(BaSO4) = 1.00*10-10

Ksp(PbSO4) = 1.70*10-8

Assume that H+ does not bind to SO42-, that ionic strength effects can be ignored, and that the system is at equilibrium.

a. Calculate the concentrations you expect to find for Pb2+, Ba2+, and SO42- in the final solution.

b. Calcualte the expected composition of the precipitate and report it as a mole fraction BaSO4 and moel fraction PbSO4 (if any is present).

answers:

a. [Pb2+] = 0.00495M, [Ba2+] = 2.91*10-5M, [SO42-] = 3.43*10-6M

b. XPbSO4 = 0.010, XBaSO4 = 0.990

Explanation / Answer

a. Looking at the Ksp value we can say that PbSO4 is more soluble in solution.

Ksp = [Pb2+][SO4^2-]

concentration of [Pb2+] = 0.01 x 0.1/0.1 + 0.101 = 4.95 x 10^-3 M = 0.00495 M

1.7 x 10^-8 = (4.95 x 10^-3)[SO4^2-]

[SO4^2-] = 3.43 x 10^-6 M

How much Ba2+ will be precipitated by this sulfate,

Ksp = 1 x 10^-10 = [Ba2+](3.42 x 10^-6)

[Ba2+] = 2.92 x 10^-5 M

b. Expected composition of precipitate

moles of PbSO4 = 0.049 x 0.201 = 9.8 x 10^-3 mols

moles of BaSO4 = 5.90 x 10^-6 mols

mole fraction PbSO4 = 0.99

moles fraction of BaSO4 = 0.01

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