The Kₛₚ for the following dissociation is 1.6 × 10 –5
PbCl₂₍ₛ₎ ⇋ Pb_((aq))²⁺ + 2Cl_((aq)) -
Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO₃)₂ and 100 mL 0.4 M NaCl ?

Solution

Chemistry ma’am · 4 steps

Dilute both ions into the 400 mL total first, then square the chloride and compare Q with Ksp.

1. Concentrations after mixing. The total volume is 300 + 100 = 400 mL. So [Pb²⁺] = (0.134 × 300)/400 = 0.1005 M and [Cl⁻] = (0.4 × 100)/400 = 0.1 M.
2. Write Q. The ionic product uses the same form as Kₛₚ. The chloride term is squared because its coefficient is 2.

   Q = [Pb²⁺][Cl⁻]² = 0.1005 × (0.1)² = 1.005 × 10⁻³

3. Compare. 1.005 × 10⁻³ is about 63 times larger than Kₛₚ = 1.6 × 10⁻⁵. When Q > Kₛₚ, the solution holds more ions than it can, so PbCl₂ precipitates.
4. So. Q > Kₛₚ, which is option (a).

The idea to keep. When two solutions are mixed, first find each ion's concentration in the total volume. Then build Q with the same powers as Kₛₚ. If Q > Kₛₚ, a precipitate forms.

Answer from the NTA final answer key. Solution by PrepWiser.