To a 250.0 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M...

To a 250.0 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNO3, 0.150M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously. The Ksp of AgBr is 5.40×10–13 and the Kf of Ag(CN)2– is 1.00×1021. What mass of AgBr would precipitate from this mixture? Do not simply write "0" as some precipitate will form.

Expert Solution

Since the complex ion has such a large Kf value (10^21), we assume that essentially all of the Ag+ from AgNO3 is complexed as Ag(CN)2-.

initial mmoles Ag+ (from AgNO3) = M Ag+ x mL Ag+ = (0.0100)(1.00) = 0.0100 mmoles Ag+
initial [Ag+] = 0.0100 mmoles Ag+ / 250 mL = 4.00 x 10^-5 M Ag+

initial mmoles CN- (from AgCN) = M CN- x mL CN- = (0.100)(1.00) = 0.100 mmoles CN-
initial [CN-] = 0.100 mmoles CN- / 250 mL = 4.00 x 10^-4 M CN-

initial mmoles Br- (from NaBr) = M Br- x mL Br- = (0.150)(1.00) = 0.150 mmoles Br-
initial [Br-] = 0.150 mmoles Br- / 250 mL = 6.00 x 10^-4 M Br-

Again, we assume that essentially all of the Ag+ is in the form Ag(CN)2-.

Molarity . . . . .Ag+ . . .+ . . .CN- . . .==> . . .Ag(CN)2-
Initial . . . . . .4 x 10^-5 . . . 4 x 10^-4 . . . . . . . . .0
Change . . . .-4 x 10^-5 . . -4 x 10^-5 . . . . . . .4 x 10^-5
Final . . . . . . . .ca. 0 . . . .3.6 x 10^-4 . . . . . .4 x 10^-5

After the above reaction, how much free Ag+ is in solution to precipitate with Br-? Reverse the Kf reaction to calculate [Ag+].

Kr = 1/Kf = 1 / (1.0 x 10^21) = 1.0 x 10^-21

Molarity . . . .Ag(CN)2- . . .==> . . .Ag+ . . .+ . . .2CN-
Initial . . . . .4 x 10^-5 . . . . . . . . . . .x . . . . . . .3.6 x 10^-4
Change . . . . . .-x . . . . . . . . . . . . . x . . . . . . . . .2x
Equilibrium .4 x 10^-5 - x . . . . . . . . x . . . . . .3.6 x 10^-4 + 2x

Kr = [Ag+][CN-]^2 / [Ag(CN)2-] = ((x)(3.6 x 10^-4 + 2x)^2) / (4 x 10^-5 - x) = 1.0 x 10^-21

Because Kr is so small, the +2x and -x terms above can be neglected, simplifying the expression to

((x)(3.6 x 10^-4)^2) / (4 x 10^-5) = 1.0 x 10^-21
(1.3 x 10^-7 x) / (4 x 10^-5) = 1.0 x 10^-21
1.3 x 10^-7 x = 4 x 10^-26
x = 3.1 x 10^-19 M = [Ag+]

AgBr will precipitate only if Qsp > Ksp.

Qsp = [Ag+][Br-] = (3.1 x 10^-19)(6.0 x 10^-4 M) = 1.86 x 10^-22 which is NOT greater than Ksp (5.4 x 10^-13).

queen_honey_blossom answered 2 years ago

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