Final answer:
The calculated reaction quotient (Q) for the formation of lead(II) bromide from a solution containing lead(II) nitrate and sodium bromide is higher than the solubility product constant (Ksp), indicating that a precipitate will form.
Step-by-step explanation:
The reaction between a solution containing lead(II) nitrate and one containing sodium bromide can lead to the formation of lead(II) bromide (PbBr2), an insoluble salt. To determine whether a precipitate will form, we compare the reaction quotient (Q) with the solubility product constant (Ksp) of PbBr2. The value of Q is calculated by multiplying the molar concentrations of the ions that would form the precipitate:
Q = [Pb2+] * [Br-]2
Given that the solution is 0.0685 M in Pb(NO3)2 and 0.0133 M in NaBr, the concentration of lead(II) ions [Pb2+] will be 0.0685 M and the concentration of bromide ions [Br-] will be 2 * 0.0133 M (due to the stoichiometry of the salt). Hence:
Q = (0.0685) * (2 * 0.0133)2 = 2.42 * 10-5
If Q < Ksp, no precipitate will form; if Q > Ksp, a precipitate will form. Since the given Ksp for PbBr2 is 4.67 * 10-6, and our calculated Q is 2.42 * 10-5, this indicates that Q > Ksp, suggesting that precipitation will indeed occur.