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 product whether a substance will precipitate or not. Precipitation is the formation of an insoluble substance. For the equation AB(s) = A^+(aq) + B^- (aq), precipitation represents a shift to the left and the production of a solid. From La chillers principle, we know that when the product of the concentrations of A^+ and B^+. This critical level, K_sp is a constant at 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. Part A- Calculate the value of Q What is the value of Q when the solution contains 2.00 times 10^-2 MSr^2+ and 1.50 times 10^-1 M CrO_4^3-? Express your answer numerically. Q = _____ Part B This question will be shown after you complete previous question(s). Part C What concentration of the barium ion Ba^2+ must be exceeded to precipitate BaF_2 from a solution that is 1.00 times 10^-2 M in the fluoride ion F? K_sp for barium fluoride is 2.45 times 10^-5 Express your answer with the appropriate units.

Explanation / Answer

Q = [Sr+2][CrO4-2]

Q = (2*10^-2)(1.5*10^-3) = 0.00003

C

Ba+2 must exceed so..

[F-] = 1*10^-2

Ksp > Q

Ksp = 1.84*10^-7

Q = [Ba+2][F-]^2

1.84*10^-7 > ([Ba+2](10^-2)^2

[Ba+2] = (1.84*10^-7)/((10^-2)^2) = 0.00184M

so it must be at least 0.00184M higher

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