Dilution Claim for a Lead Iodide Precipitate
A student mixes 20.0 mL of 0.100 M Pb(NO3)2 with 20.0 mL of 0.100 M KI. A yellow precipitate forms immediately. The student does not filter it, then adds 10.0 mL of distilled water to the mixture and stirs. The student claims, 'The precipitate will dissolve because dilution lowers the concentration of Pb2+ and I−, so the reaction quotient Q becomes smaller than Ksp.' The student has a table value Ksp(PbI2) = 7.1 × 10−9 at 25 °C. The stock solutions are at 25 °C, and volumes are additive. The student wants to know whether the precipitate will disappear completely, partially dissolve, or remain essentially unchanged.
1. Determine the limiting reagent and the composition of the mixture after precipitation but before any dissolution. Include mmol of precipitate and excess ion.
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Initial moles: Pb2+ = 0.100 mol/L × 0.0200 L = 2.00 mmol; I− = 2.00 mmol. The precipitation reaction is Pb2+ + 2I− → PbI2(s). I− is limiting: 2.00 mmol I− consumes 1.00 mmol Pb2+ and forms 1.00 mmol PbI2. After precipitation, about 1.00 mmol Pb2+ remains in excess, I− is essentially consumed, and 1.00 mmol solid PbI2 is present. The total volume before the water addition is 40.0 mL, so [Pb2+] is about 1.00 mmol/40.0 mL = 0.025 M if all remaining Pb2+ is dissolved.
2. Evaluate the dilution claim quantitatively: calculate the equilibrium iodide concentration after adding 10.0 mL water and the percentage of precipitate that dissolves.
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After adding 10.0 mL water, total volume is 50.0 mL and [Pb2+] = 1.00 mmol/50.0 mL = 0.0200 M. Let x be the molar solubility of PbI2 in this solution. Dissolution gives [I−]=2x and [Pb2+]≈0.0200+ x ≈0.0200 M because x is small. Ksp = [Pb2+][I−]^2 ≈ 0.0200(2x)^2 = 0.0800x^2 = 7.1×10−9. Thus x^2=8.9×10−8, x=3.0×10−4 M. Dissolved PbI2 = x × 0.0500 L = 1.5×10−5 mol, compared with 1.0×10−3 mol solid, about 1.5%. The precipitate does not disappear; it partially dissolves by a tiny amount. The student's claim is false because the excess Pb2+ common ion keeps Q at Ksp with most solid still present.
3. Would adding 10.0 mL of 0.100 M KI instead of water dissolve more, less, or the same amount? Explain using Q or Le Châtelier's principle.
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Adding 10.0 mL of 0.100 M KI adds 1.00 mmol I−. That I− reacts with the 1.00 mmol excess Pb2+: 1.00 mmol I− consumes 0.50 mmol Pb2+ and forms 0.50 mmol more PbI2, leaving 0.50 mmol Pb2+ in excess. Thus more precipitate forms, and less solid dissolves than with pure water. In equilibrium terms, the added I− makes Q=[Pb2+][I−]^2 larger than Ksp, so precipitation occurs until Q returns to Ksp. The common ion shifts the solubility equilibrium toward the solid, not toward dissolution.