A concentrated aqueous solution of Pb(NO3)2 is slowly added to 1.0 L of a mixed aqueous solution containing 0.010 M Na2CrO4 and 2.5 M NaBr in an attempt to selectively precipitate only one of the anions. The sparingly soluble salt that precipitates first (neglecting any volume change) is with a [Pb2+] concentration of .

Ksp(PbCrO4) = 1.8 × 10–14, Ksp(PbBr2) = 6.3 × 10–6

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

Explanation:

Step 1: Data given

Molarity of Na2CrO4 = 0.010 M

Molarity of NaBr = 2.5 M

Ksp(PbCrO4) = 1.8 * 10^–14

Ksp(PbBr2) = 6.3 * 10^–6

Step 2: The balanced equation

PbCrO4 →Pb^2+ + CrO4^2-

PbBr2  → Pb^2+ + 2Br-

Step 3: Define Ksp

Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]

1.8*10^-14 = [Pb^2+] * 0.010 M

[Pb^2+] = 1.8*10^-14 /0.010

[Pb^2+] = 1.8*10^-12 M

The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M

Ksp PbBr2 = [Pb^2+][Br-]²

6.3 * 10^–6 = [Pb^2+] (2.5)²

[Pb^2+] = 1*10^-6 M

The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

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