Question

Calculate the solubility at 25 °C of PbCrO4 in pure water and in a 0.0070 M Na, CrO 4 solution. You'll find K, data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: FF solubility in 0.0070 M Na2CrO4 solution: x 6 ?

Answer 1

The dissolution of PbCrO4(s) in aqueous solution takes place according to the following equation

               PbCrO4(s) + H2O(aq) <-----------> Pb2+(aq) + CrO42-(aq)

The solubility product constant of the above reaction is given by

Ksp = [Pb2+] [CrO42-] = 1.8*10^-14

BUT [Ca2+] = [CO32-] = s moles/L

Ksp = s*s = 4.5*10^-9

s = sqrt(4.5*10^-) = 1.34*10^-7 M or 1.34*10^-7 moles/litre.

i.e 1.34*10^-7 moles lead chromate are dissolved in 1 Liter of the solution

ANS : 4.3*10^-5 grams of lead chromate is dissolved in 1 Liter of the solution

Upon the addition of 0.0070 M NaCrO4, the solubility product constant of the above reaction is given by

Ksp = [Pb2+] [CrO42-] = [Pb2+] *0.0070 = 1.8*10^-14

[Pb2+] = 2.57*10^-12 M OR 2.57*10^-12 moles/liter

i.ei.e 2.57*10^-12 moles are lead chromate dissolved in 1 Liter of the solution

   8.3*10^-10 grams of lead chromate is dissolved in 1 Liter of the solution in the presence of 0.0070 M Na2CrO4

SOLUBILITY IN PURE WATER : 4.3*10^-5 grams/liter

SOLUBILITY IN 0.0070 M Na2CrO4 solution : 8.3*10^-10 grams/liter