A gallon jug of water has 3.79 L H_2 K_sp for PbS(s) is 9.04 times 10^-29. Write

Question

A gallon jug of water has 3.79 L H_2 K_sp for PbS(s) is 9.04 times 10^-29. Write the dissolution reaction for PbS(s), assuming the lead is +2. How many grams of pb^2+ will be found in that jug? Assume no other ions are present. Given the K_sp for this reaction, what is DG degree for the dissolution reaction at equilibrium? Assume 25 (not 25.0) degrees Celsius. A water sample at the same temperature has 1.23 times 10^-10 M of Pb^2+ and 4.56 times 10^-10 M of S^2-. What is DG for the dissolution reaction? Will more lead dissolve, will it precipitate, or neither? (If you didn't answer the above question, use DG degree = 1 234 kJ/mol.)

Explanation / Answer

5)

dissociation equation :

PbS (s) -----------------------------> Pb+2 (aq) + SO4-2 (aq)

Ksp = [Pb+2] [SO4-2]

9.04 x 10^-29 = S x S

S = 9.51 x 10^-15 M

moles of Pb+2 = 9.51 x 10^-15 x 3.79

                        = 3.60 x 10^-14

mass of Pb+2 = 3.60 x 10^-14 x 207.2

                        = 7.47 x 10^-12 g

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