Percent Yield in a Lead(II) Iodide Precipitation
During a precipitation lab, a group is asked to prepare lead(II) iodide and report a percent yield. They pipet 25.0 mL of 0.200 M Pb(NO3)2 into a beaker and add 30.0 mL of 0.150 M KI with stirring; a yellow solid forms at once. The mixture is filtered on a Büchner funnel, the solid rinsed twice with 5 mL portions of distilled water, and the funnel dried at 110 °C for 20 minutes in an oven shared with a group that had been drying a hydrated salt. The cooled solid weighs 0.86 g. The filtrate still shows a faint yellow tint after the first rinse. One student wants to know why adding more KI would not simply "make more product," and another asks whether the 0.86 g figure can be trusted. The instructor reminds them that lead(II) iodide is slightly soluble and that iodide ion can also act as a ligand. Molar mass of PbI2 = 461.0 g/mol.
1. Identify the limiting reactant, calculate the theoretical yield of PbI2, and compute the group's percent yield.
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Convert both reactants to moles before comparing anything:
- n(Pb2+) = 0.0250 L × 0.200 mol/L = 5.00 × 10^-3 mol
- n(I-) = 0.0300 L × 0.150 mol/L = 4.50 × 10^-3 mol
The balanced precipitation is Pb2+ + 2 I- → PbI2(s), so each mole of Pb2+ requires two moles of iodide. The available Pb2+ would demand 1.00 × 10^-2 mol I-, and only 4.50 × 10^-3 mol is present. Iodide is therefore limiting and lead(II) ion is in excess: 2.25 × 10^-3 mol Pb2+ is consumed and 2.75 × 10^-3 mol remains in roughly 55 mL of solution.
Mass of product from the limiting reactant: 4.50 × 10^-3 mol I- ÷ 2 = 2.25 × 10^-3 mol PbI2; 2.25 × 10^-3 mol × 461.0 g/mol = 1.037 g.
Percent yield = 0.86 g ÷ 1.037 g × 100% = 83%.
A useful self-check: the excess Pb2+ is the common ion that suppresses dissolution of the precipitate, so solubility alone cannot be the whole 17% deficit. Naming the limiting reactant is not the answer to the question; the reasoning that iodide runs out first is.
2. The group's mass is below the theoretical value. Evaluate the plausible contributions to the deficit and to any possible overestimate, and decide which explanation is best supported by the observations.
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Losses push the mass down; contamination and incomplete drying push it up. Working through them:
- Solubility. PbI2 has Ksp ≈ 7.1 × 10^-9, giving a solubility near 1.2 × 10^-3 M in pure water, about 0.03 g in 55 mL of filtrate. Excess Pb2+ (about 0.05 M) suppresses this by the common-ion effect, so dissolution is real but small.
- Transfer and filtration losses. Fine precipitate passing the filter paper or clinging to the beaker commonly costs a few percent. The faint yellow tint in the filtrate after the first rinse supports this explanation directly.
- Incomplete drying would raise the mass, not lower it, and a shared oven used for a hydrate risks contamination that could push the result above 100%.
Because the result is 83% rather than above 100%, wetness and contamination are not the dominant issues. The best-supported explanation is a combination of product lost through the filter and residual solubility, with the tinted filtrate as the strongest single piece of evidence. A defensible improvement is to use a finer filter, allow digestion before filtering, and rinse with a dilute Pb(NO3)2 solution rather than pure water.
3. A classmate proposes adding 60.0 mL of 0.150 M KI instead of 30.0 mL, arguing the yield must increase. Predict and justify what would actually be observed, and state the general principle being tested.
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Adding iodide up to the stoichiometric point does increase the amount of precipitate, because iodide remains the limiting reagent and each added mole produces half a mole of PbI2 until 1.00 × 10^-2 mol I- is present (66.7 mL of 0.150 M KI). Doubling the iodide to 60.0 mL gives 9.00 × 10^-3 mol I-, still below the 1.00 × 10^-2 mol needed to consume all the lead, so the yield of precipitate would rise sharply, to 4.50 × 10^-3 mol PbI2, or 2.07 g.
Past that equivalence point the reasoning breaks down. Lead(II) forms soluble iodoplumbate complexes, PbI2(s) + 2 I- ⇌ [PbI4]2-, so a large iodide excess redissolves precipitate rather than creating more. Percent yield would fall, even though the theoretical yield based on lead is fixed at 5.00 × 10^-3 mol Pb2+ → 2.31 g PbI2.
The principle under test is that the mass of isolated product is not a monotonic function of the amount of any one reagent. Maximum recovery requires a modest excess of one reagent, not an unlimited excess, and the choice of which reagent to keep in excess must consider complexation, not only solubility and the balanced equation.