The next three problems refer to the following fractional precipitation experime

Question

The next three problems refer to the following fractional precipitation experiment. Suppose a solution contains 0.6 M Ba^2+ (aq) and 0.8 M Mg^2+(aq). Fluoride ions, F^-, are then slowly added to the solution. At what [F^-] will BaF_2(s) start to precipitate? (Suggestion: Write the dissolving equation and K_sp expression for BaF_2 before starting this problem.) 3.0 times 10^-7 M 4.7 times 10^-4 M 2.7 times 10^-4 M 9.5 times 10^-4 M 5.5 times 10^-4 M At what [F^-] will MgF_2(s) start to precipitate? (Suggestion: Write the dissolving equation and K_sp expression for MgF_2 before starting this problem.) 8.8 times 10^-11 M 1.1 times 10^-5 M 9.4 times 10^-6 M 4.7 times l0^-6 M 1.9 times 10^-5 M Calculate the [Mg^2+] left in solution when Ba^2+ starts to precipitate, and then answer the following question. What percent of the original Mg^2+ is left in solution when Ba^2+ starts to precipitate? 1.6 times 10^-5 % 1.2 times 10^-1 % 1.6 times 10^-7 % 2.3 times 10^-2 % 2.9 times 10^-2 %

Explanation / Answer

Q24

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

so..

[Ba+2] = 0.6 M

[F-] = sqrt(Ksp/([Ba+2])) = sqrt(1.84*10^-7)/(0.6))

[F-] = 0.0007149 M = 7.2*10^-4; nearest answer is d (use your Ksp value)

Q25

MgF2 <> Mg+2 + 2F-

Ksp = [Mg+2][F-]^2

[F-] = sqrt(Ksp)/([Mg+2])

[F-] = sqrt((5.16*10^-11)/(0.8)) = 0.00000803118 = 8*10^-6; nearest answer i C

Q26

[F-] = 7.2*10^-4 when Ba preicpitates

so...

Ksp = [Mg+2]F-]

5.16*10^-11 = [Mg+2](7.2*10^-4)

[Mg+2] = (5.16*10^-11)/(7.2*10^-4) = 7.166666*10^-8

so..

% = 7.166666*10^-8 / 0.8 * 100 = 8.9583*10^-5 % ; nearest is a

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