# 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)

###### 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:

$\mathbf{Q}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{Q}\mathbf{=}\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{Br}}^{-}\right]}^{\mathbf{2}}}$

Concentration of Pb2+:

Pb(NO3)2(aq)  Pb2+(aq) + 2 NO3(aq)

Given: 0.0600 M in Pb(NO3)2

87% (11 ratings) ###### 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)