Question

A solution containing Pb(NO₃)₂ is mixed with one containing NaBr to form a solution that is 0.0150 M Pb(NO₃)₂ and 0.0350 M in NaBr. Does a precipitate form in this solution? Ksp = 4.67 x 10⁻⁶ = [Pb²⁺][Br]²

a. Yes
b. No

Answer 1

When mixing solutions of Pb(NO₃)₂ and NaBr, the resulting ion product (1.8375 × 10⁻µ) is greater than the solubility product constant (Ksp = 4.67 × 10⁻⁶) for PbBr₂, indicating that a precipitate of PbBr₂ will indeed form.

Step-by-step explanation:

To determine whether a precipitate will form when solutions of Pb(NO₃)₂ and NaBr are mixed, we must compare the ion product to the solution's Ksp. For lead (II) bromide (PbBr₂), the reaction would be:

Pb(NO₃)₂ (aq) + 2 NaBr (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)

To calculate the ion product (Q) for PbBr₂, we use the concentrations of Pb²⁺ and Br⁻ ions in the mixed solution:

Q = [Pb²⁺][Br⁻]² = (0.0150 M)(0.0350 M)² = 1.8375 × 10⁻µ

Since Q = 1.8375 × 10⁻µ is greater than the given Ksp of 4.67 × 10⁻⁶ for PbBr₂, a precipitate of PbBr₂ will form.