1) A solution contains 5.27×10-3 M chromium(III) acetate and 5.78×10-3 M zinc ni

Question

1) A solution contains 5.27×10-3 M chromium(III) acetate and 5.78×10-3 M zinc nitrate. Solid potassium phosphate is added slowly to this mixture. What is the concentration of chromium(III) ion when zinc ion begins to precipitate? [Cr3+] = M

2) A solution contains 1.35×10-2 M barium acetate and 9.52×10-3 M lead nitrate. Solid sodium sulfate is added slowly to this mixture. What is the concentration of barium ion when lead ion begins to precipitate? [Ba2+] = M

3 )A solution contains 1.07×10-2 M potassium chloride and 1.49×10-2 M sodium sulfate. Solid lead nitrate is added slowly to this mixture. What is the concentration of sulfate ion when chloride ion begins to precipitate? [sulfate] = M

Explanation / Answer

1) Need the Ksp values for CrPO4 (chromium phosphate) and Zn3(PO4)2 (zinc phosphate) since these values aren’t easily available off the internet.

2) The salts formed are barium sulfate, BaSO4 (Ksp = 1.08*10-10) and lead sulfate, PbSO4 (Ksp = 2.53*10-8). Since both BaSO4 and PbSO4 are both 1:1 salts, we can simply compare the Ksp values to say that PbSO4 is more soluble (higher Ksp) than BaSO4. Hence, BaSO4 will precipitate out first followed by PbSO4.

We need to determine [SO42-] when PbSO4 begins to precipitate. Write down the Ksp expression for PbSO4 as below.

PbSO4 (s) <======> Pb2+ (aq) + SO42- (aq)

Ksp = [Pb2+][SO42-]

=====> 2.53*10-8 = (9.52*10-3)*[SO42-]

=====> [SO42-] = (2.53*10-8)/(9.52*10-3) = 2.66*10-6 M.

We need to determine [Ba2+] when [SO42-] = 2.66*10-6 M. Write down the ionization of BaSO4 as below.

BaSO4 (s) <======> Ba2+ (aq) + SO42- (aq)

Ksp = [Ba2+][SO42-]

=====> 1.08*10-10 = [Ba2+]*(2.66*10-6)

=====> [Ba2+] = (1.08*10-10)/(2.66*10-6) = 4.06*10-5 M (ans).

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