What is the maximum volume of 8.6 × 10^-5 M K₂CrO₄ that,added to 40.4 mL of a solution that is 5.1 × 10^-5 M Ba(NO₃)₂ and 1.5 × 10^-6 M Pb(NO₃)₂,will precipitate PbCrO₄ but not BaCrO₄? For PbCrO₄,K_sp = 1.8 × 10^-14,and for BaCrO₄,K_sp = 1.2 × 10^-10.

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the precipitation reaction: Pb(NO₃)₂ + K2CrO₄ → PbCrO₄ + 2K(NO₃) The equilibrium expression for this reaction is as follows: K_sp = [PbCrO₄][K(NO₃)]2/[Pb(NO₃)][K2CrO₄] We want to find the maximum volume of the potassium chromate solution that can be added to the lead nitrate and barium nitrate solution without precipitating barium chromate. This means we need to find the equilibrium concentration of Ba2+ and compare it to the Ksp of BaCrO4. First, let's find the concentration of Pb2+ and NO3- in the initial solution: [Pb2+] = 1.5 × 10^-6 M [NO3-] = 1.5 × 10^-6 M (since Pb(NO₃) is a strong electrolyte and dissociates completely) Next, we need to find the amount of Ba(NO3)2 that will react with the added K2CrO4. Since the stoichiometric ratio of K2CrO4 to Ba(NO3)2 is 1:1, the amount of Ba(NO3)2 that reacts is: n(Ba(NO3)2) = [Ba2+]initial × Vinitial where Vinitial is the volume of the initial solution (40.4 mL). Note that we assume that the volume of the added K2CrO4 solution is negligible compared to the initial volume. We can find [Ba2+]initial using the initial concentration and the volume of the initial solution: [Ba2+]initial = 5.1 × 10^-5 M n(Ba(NO3)2) = [Ba2+]initial × Vinitial = (5.1 × 10^-5 M) × (40.4 mL/1000 mL) = 2.0604 × 10^-4 mol Since the stoichiometric ratio of Ba(NO3)2 to BaCrO4 is 1:1, the amount of BaCrO4 that will precipitate is also 2.0604 × 10^-4 mol. Now we can set up an ice table to find the equilibrium concentration of Pb2+ and NO3- after the reaction: Pb(NO₃)₂ + K2CrO₄ → PbCrO₄ + 2K(NO₃) I 1.5 × 10^-6 M 1.5 × 10^-6 M 0 0 C 1.5 × 10^-6 M - x 1.5 × 10^-6 M - x 2x 2x E 1.5 × 10^-6 M - x 1.5 × 10^-6 M - x 2x 2x The equilibrium concentrations of Pb2+ and NO3- are therefore: [Pb2+] = x = 1.8 × 10^-14 M [NO3-] = 1.5 × 10^-6 M - x = 1.5 × 10^-6 M Now we can find the maximum volume of the K2CrO4 solution that can be added without precipitating BaCrO4: Ksp(BaCrO4) = [Ba2+][CrO42-] 1.2 × 10^-10 = x × (2x)2/(2.0604 × 10^-4 - x) Solving for x, we get x = 1.066 mL. Therefore, the maximum volume of the K2CrO4 solution that can be added is: V = final volume - initial volume = (1.066 mL + 40.4 mL)/1000 mL = 0.04146 L V = 1.1 mL (rounded to two significant figures) Therefore, the correct answer is B.

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