Chemistry 102: Acid Equilibrium: Calculate Q and answer multiple choice question

Question

9). At 25^oC the K_sp for PbCl₂ is 1.6 × 10^–5.

part 1). Calculate Q  for the following: 125.0 mL of 0.0900 M Pb(NO₃)₂ is mixed with 75.0 mL of 0.0200 M NaCl at 25^oC

part 2). Will a precipitate form from the above reaction?

Choose one:

A) Yes

B) No

C) not enough info to answer the question

Answer 1

Q is the value of the K expression with the initial concentrations input. (It is often called R for solubility equilibria)

The K_sp = 1.6 x 10^-5 is for the reaction PbCl₂ ↔ Pb²⁺ + 2Cl^- which has the K_sp expression, [Pb²⁺][Cl^-]².

We need the concentrations of Pb²⁺ and Cl^- given the dilution after mixing: M' = M(V/V')

[Pb²⁺] = .09M(125 ml/200ml) = .05625 M

[Cl^-] = .02M(75 ml/200 ml) = .0075 M

The criteria for the Q is standard: If Q > K_sp the backward reaction is favored (ppt'n), if Q < K_sp the forward reaction is favored (dissociation), and if Q = K_sp the reaction is in equilibrium (sat'n)

In this case (.05625)(.0075)² = 3.16 x 10^-6 < K_sp

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Answer 2

1).

moles Pb(NO3)2 = 125.0 ml x 1 L / 1000 ml x 0.0900 mol / L = 0.01125 mols

mols NaCl = 75.0 ml x 1 L / 1000 ml x 0.0200 mol / L =1.50x10^-3 mols

Total volume = 125.0 ml + 75.0 ml = 200.0 ml = 0.200 L

[Pb²⁺] = 0.01125 mol / 0.200 L = 0.05625 M

[Cl^-] = 1.50x10^-3 mol / 0.200 L = 7.5x10^-3 M

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

Q = [Pb²⁺][Cl^-]² = (0.05625)(7.5x10^-3)² = 3.16x10^-6

2). No, a precipitate will not form. The reason for this is that the value of Q is less than the Ksp value. This tells us that the concentrations of the Pb²⁺ and Cl^- are not high enough to cause a precipitate to form.