(please explain as well as answer :)) 11) Lead iodide, a sparingly soluble salt,

Question

(please explain as well as answer :)) 11) Lead iodide, a sparingly soluble salt, dissolves in solution according to the following equation. PbI2(s) Pb2+(aq) + 2 I-(aq) Ksp = 7.1 x 10-9 Enough PbI2 is added to create a saturated liter of aqueous solution. Suppose that 6.00 g of Pb(NO3)2 (a soluble salt, Pb(NO3)2(s) Pb2+(aq) + 2 NO3-(aq)) is added to the solution (assume the volume does not change). What would you expect to change as a result of the addition of Pb(NO3)2 ? A) More PbI2 will dissolve B) The Ksp will decrease C) The Ksp will increase E) The solubility of PbI2 will decrease D) The pH of the solution will decrease b) calculate the molar solubility of PbI2 after the addition of 6.00 g of Pb(NO3)2 to the saturated PbI2 solution. Answer _____________________ Comment

Explanation / Answer

PbI2(s) Pb2+(aq) + 2 I-(aq) ; Ksp = 7.1 * 10^-9
number of moles of Pb(NO3)2 = 6/331.2 = 0.0181
concentration of Pb(NO3)2 = 0.0181/1 = 0.0181 mol/L
Pb(NO3)2 is strong electrolyte means it is completely dissociated into its ions then,
concnetration of Pb+2 ion = 0.0181 mol/L
If you add Pb(NO3)2 to PbI solution then the concentration of Pb+2 ions increases and then the equlibrium shift to left side which indicates more amount of PbI2 will be formed
So Solubility of PbI2 will decrease.
Ksp = [Pb+2][I-]^2
Ksp = (s)*(s)^2
after adding Pb(NO3)2 to PbI2 then solubility S = (s+c) where C = concentration of Pb+2 ion from Pb(NO3)2)
S = (S + 0.0181)
Ksp = (s + 0.0181)*(s)^2
7.1 * 10^-9 = (s + 0.0181)*(s)^2
S = 6.1592 * 10^-4

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