The equilibrium constant for the reaction of lead sulfate dissolving in water is 1.5 × 10-8. If enough PbSO4 is added for equilibrium to be reached, what will be the equilibrium concentration of Pb2+? PbSO4 (s) Pb2+ (aq) + SO42- (aq)

Question

The equilibrium constant for the reaction of lead sulfate dissolving in water is 1.5 × 10-8. If enough PbSO4 is added for equilibrium to be reached, what will be the equilibrium concentration of Pb2+?

PbSO4 (s) <-> Pb2+ (aq) + SO42- (aq)

sweetburger · Answer

Been a while since I've done these questions someone might need to correct me.

Keq = [Pb^2+][SO4^2-]
PbSO4 is not important as it is a solid and it will not affect the equation

we know Keq= 1.5x10^-8

1.5x10^-8 = [x][x]

1.5x10^-8 = x^2

square both sides and use the positive x value for your answer.

x is equivalent to both Pb^2+ and SO4^2+ so you have found the concentration of both with one calculation.

sweetburger · Answer

*square root both sides.