A solution of NaF is added dropwise to a solution that is 0.0144 mol L^-1 in Ba²⁺. When the concentration of F^- exceeds ________ mol L^-1, BaF₂ will precipitate. Neglect volume changes. For BaF₂, K_sp = 1.7 × 10^-6. A)5.9 × 10^-5 B)1.1 × 10^-2 C)2.4 × 10^-8 D)2.7 × 10^-3 E)1.2 × 10^-4

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Quizplus · Accepted Answer

First, we need to write out the balanced equation for the reaction: Ba²⁺ + 2NaF^- → BaF₂ + Na2^- We can see that we are adding NaF to a solution containing Ba²⁺. The question asks us to find the concentration of F^- needed to precipitate BaF₂. This means that we need to find the concentration at which Qsp (the reaction quotient) equals Ksp, the solubility product constant for BaF₂. At this point, the solution is saturated with BaF₂ and any additional F^- will result in precipitation. The expression for Qsp is: Qsp = [BaF₂][Na2^-]^2 We can assume that the initial concentration of Ba²⁺ is much larger than the amount of NaF added, so we can neglect the change in its concentration. At the beginning of the reaction, there is no BaF₂ precipitate, so we can ignore its concentration in the Qsp expression. That leaves us with: Qsp = [Na2^-]^2[Ksp] Substituting in the values given in the question: Ksp = 1.7 × 10^-6 [NaF^-] = 0.0144 mol L^-1 [Na2^-] = 2 × [NaF^-] = 2 × 0.0144 mol L^-1 = 0.0288 mol L^-1 Plugging in these values: Qsp = (0.0288 mol L^-1)^2 × (1.7 × 10^-6) = 1.39 × 10^-9 Now we just need to find the concentration of NaF^- at which Qsp equals Ksp: Ksp = Qsp [Na2^-]^2[Ksp] = (0.0288 mol L^-1)^2 × (1.7 × 10^-6) [Na2^-]^2 = 1.39 × 10^-3 [Na2^-] = sqrt(1.39 × 10^-3) = 1.18 × 10^-7 Remember that [Na2^-] is twice [NaF^-], so: [NaF^-] = 0.59 × 10^-7 = 5.9 × 10^-6 Therefore, the answer is B) 1.1 × 10^-2.

A Solution of NaF Is Added Dropwise to a Solution