To a 150.0mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNO3,...

To a 150.0mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNO_3, 0.225 M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously. What mass of AgBr would precipitate from this mixture? (Hint: The K_sp of AgBr is 5.4x10^-13 and the K_f of Ag(CN)₂^- is 1.0x10²¹)

Expert Solution

AgBr(s )-------- Ag+(aq) + Br–(aq) Ksp=[Ag][Br]= 5.4*10^-13

Ag+(aq) + 2CN–(aq) -------- Ag(CN)2–(aq) Kf=[Ag(CN)2]/[Ag]*[CN]^2 = 1.0* 10²¹

so the net equation would be
AgBr(s) + 2CN–(aq) ------ Ag(CN)2–(aq) + Br–(aq)

the initial concentration of CN- was not 0.100M. The problem states that 1. mL of 0.100M NaCN was added to a 150 mL volumetric flask and then the other two solutions and distilled water were added until the mark. That

K_f of Ag(CN)₂^- is 1.0x10²¹ has such a large Kf value , we assume that essentially all of the Ag+ from AgNO3 is complexed as Ag(CN)2-.

find initial mmoles of Ag+ (from AgNO3) = M Ag+ x mL Ag+ = (0.0100)(1.00) = 0.0100 mmoles Ag+
initial [Ag+] = 0.0100 mmoles Ag+ / 150 mL = 6.66 x 10^-5 M Ag+

find initial mmoles CN- (from NaCN) = M CN- x mL CN- = (0.100)(1.00) = 0.100 mmoles CN-
initial [CN-] = 0.100 mmoles CN- / 150 mL = 6.66 x 10^-4 M CN-

find initial mmoles Br- (from NaBr) = M Br- x mL Br- = (0.225)(1.00) = 0.225 mmoles Br-
initial [Br-] = 0.225 mmoles Br- / 150 mL = 1.5 x 10^-3 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 . . . . . .6.66 x 10^-5 . . . . . 6.66 x 10^-4 . . . . . . . . .0
Change . . .6.66 x 10^-5 .. . . .. . 6.66 x 10^-5 . . . . . . .6.66 x 10^-5
Final . . . . . . . . 0 . . . . . . . . . . . . .5.99 x 10^-4 . . . . . .. .6.66 x 10^-5

After the above reaction, how much free Ag+ is in solution to precipitate with Br- and 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 . . . . .6.66 x 10^-5 . . . . . . . . .x . . . . . . .5.99x 10^-4
Change . . . . . .-x . . . . . . . . . . . . . . .. x . . . . . . . . .2x
Equilibrium . 6.66 x 10^-5 - x . . . . x . . . . . .5.99 x 10^-4 + 2x

Kr = [Ag+][CN-]² / [Ag(CN)^2-] = ((x)(5.99 x 10^-4 + 2x)²) / (6.66 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)(5.99 x 10^-4)²) / (6.66 x 10^-5) = 1.0 x 10^-21
(1.3 x 10^-7 x) / (4 x 10^-5) = 1.0 x 10^-21
x = 1.856 x 10^-19 M = [Ag+]

AgBr will precipitate only if Qsp > Ksp.

Qsp = [Ag+][Br-] = (1.856 x 10^-19)(1.5 x 10^-3 M) = 2.784 x 10^-22 which is NOT greater than Ksp (5.4 x 10^-13).

then AgBr not precipitate

queen_honey_blossom answered 1 year ago

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