5. The following questions relate to Pbl2, whose Ksp is 7.1x10 9. a. Use the Ksp

Question

5. The following questions relate to Pbl2, whose Ksp is 7.1x10 9. a. Use the Ksp value given above to find how many grams of lead (II) iodide can be dissolved in 1.00 L of water. b. You have a 1.00 L solution that has a lead (II) ion concentration of 0.0015 M. Will lead (II) iodide precipitate if 0.30 g of Nal is added to the solution? Show your answer by a calculation. No credit given for a guess. Assume the total volume is unchanged. -28 c. The Ksp for lead (II) sulfide is 3.0x10%. If a solution contains equal concentrations of sulfide and iodide ions and Pb(NO3)2 is added, which solid will precipitate first Pbl2, or PbS? Explain your answer.

Explanation / Answer

a)

Ksp = [Pb+2][I-]^2

substitute known data, assume

[PbI2] = [Pb+2] = 2*[I-]

then

Ksp = (S)(2S)^2

7.1*10^-9 = (4*S^3)

S = ((7.1*10^-9)/4)^(1/3)

S = 0.00121 mol per liter of PbI2

b)

[Pb+2] = 0.0015 M

[I-] = [NaI] = mass/MW / V = (0.3/149.89424)/(1) = 0.00200

If

Q >Ksp, then precipitate forms

Q = [Pb+2][I-]^2 = (0.0015 )(0.00200^2) = 6*10^-9

note that

6*10^-9 < Ksp, therefore, there will be No precipitate

c)

PbS ; Ksp = 3*10^-28

Solubility of [Pb+2] required

(3*10^-28) = [Pb+2] ( S-2)

[S-2] = (3*10^-28) / [Pb+2]

For I-

[I-] = sqrt((7.1*10^-9) / [Pb+2] )

clearly

S-2 forms PbS faster, since it is less soluble

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