Problem: A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0150 M in Pb(NO3)2 and 0.00350 M in NaBr. If the solutions are concentrated through evaporation and mixed again to form a solution that is 0.0600 M in Pb(NO3)2 and 0.0158 M in NaBr, will a precipitate form in this newly mixed solution? (The value of Ksp for PbBr2 is 4.67 x 10–6)

🤓 Based on our data, we think this question is relevant for Professor Wambsgans' class at DREXEL.

FREE Expert Solution

For this, we need to compare the reaction quotient (Q) vs. the solubility product constant (Ksp)

Recall that when:

• Q > Ksp: the solution is supersaturated and a precipitate will form. Reactants are favored.

• Q = Ksp: the solution is at equilibrium and no precipitate will form.

• Q < Ksp: the solution is unsaturated and no precipitate will form. Products are favored.


The expected precipitate is PbBr2. The bromide ion, Br, has a charge of –1. Pb then has a charge of +2. 

The dissociation of PbBr2 in water is as follows:


PbBr2(s)  Pb2+(aq) + 2 Br(aq)            Ksp = 4.67x10–6


The reaction quotient expression for PbBr2 is:

Q=productsreactantsQ=[Pb2+][Br-]2



Concentration of Pb2+:

Pb(NO3)2(aq)  Pb2+(aq) + 2 NO3(aq)

Given: 0.0600 M in Pb(NO3)2 

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Problem Details

A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0150 M in Pb(NO3)2 and 0.00350 M in NaBr. If the solutions are concentrated through evaporation and mixed again to form a solution that is 0.0600 M in Pb(NO3)2 and 0.0158 M in NaBr, will a precipitate form in this newly mixed solution? (The value of Ksp for PbBr2 is 4.67 x 106)