To understand the relationship between precipitation and the solubility product

Question

To understand the relationship between precipitation and the solubility product and to be able to predict whether a substance will precipitate or not. Precipitation is the formation of an insoluble substance. For the equation AB(s) doubleheadarrow A^+ (aq) + B^- (aq), precipitation represents a shift to the left and the production of a solid. From lessthanorequalto Chatelier's principle, we know that when the product of the concentrations of A^+ and B^- gets above a certain level, the reaction will respond by shifting left to decrease the concentrations of A^+ and B^-. This critical level, K_sp, is a constant at a certain temperature. In this case. Q = [A^]+ [B^-]. where Q is the ion product. When Q > K_sp, precipitation occurs until Q is equal to K_sp When Q lessthanorequalto K_sp, a precipitate does not form. What is the value of Q when the solution contains 2.00 Times 10^-3 M Ca^2+ and 3.00 Times 10^-2 MSO^2-_4? Express your answer numerically. Q = What concentration of the lead ion. Pb^2+, must be exceeded lo precipitate PbCl_2 from a solution that is 1.00 Times 10^-2 M in the chloride ion. Cl^-? K_sp for lead(ll) chloride is 1.17Times 10^-5. Express your answer with the appropriate units.

Explanation / Answer

A)

Q = [procuts]^p / [reactats]^r

Q = [Ca+2][SO4-2] / 1

since CaSO4(s) has activity = 1,

Q = (2*10^-3) * (3*10^-2) = 0.00006

C)

Find:

Pb+2 + 2Cl- <--> PbCl2(s)

recall that, if Q > Ksp, there will be precipitation

so..

Q = [Pb+2][Cl-]^2

so

Q = (10^-2)^2 * [Pb+2]

[Pb+2] = (1.17*10^-5) / ((10^-2)^2 ) = 0.117 M of Pb+2

so, this will be in equilibrium

anything more will start precipitation

so

[Pb+2] > 0.117 M

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