to a 100.0 mL volumetric flask are added 1.00 mL volumes of three solutions: .0100 M...

to a 100.0 mL volumetric flask are added 1.00 mL volumes of three solutions: .0100 M AgNO3, .205 M NaBr, and .100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigerously. What mass of AgBr would precipitate from this mixture? Hint: The Ksp of AgBr is 5.4 x 10-13 and the Kf of Ag(CN)2- is 1.0 x 10^21

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Expert Solution

Moles of AgNO3 (OR moles of Ag+ ) = Molarity X Volume = 0.01 X 1mL = 0.01mmoles

Molarity of AgNO3 when the solution is diluted to 100mL = 0.01mmol / 100mL = 1.0 X 10^-4M

Similarly Moles of NaBr (Or moles of Br-) = 0.205 X 1mL = 0.205mmol

Molarity = 0.205mmol/100mL = 2.05 X 10^-3M

Moles of NaCN = 0.1mmol

Molarity = 0.1mmol / 100mL = 1.0 X 10^-3 M

Since Kf is very large hence almost all the Ag+ will be in complex form Ag(CN)2-

Ag+ will react with CN-

Ag+ + 2CN^- -------> Ag(CN)₂^-

I 1.0 X 10^-4M ---1.0 X 10^-3M --------- 0

C -1.0 X 10^-4M --- -2X1.0 X 10^-4M -------- 1.0 X 10^-4M

E 0 ------- 8 X 10^-4 --------1.0 X 10^-4M

From above ICE equation it is obvious that all the Ag+ is consumed in Ag(CN)2- formation. The only Ag+ available will be from dissiciation of the cyanide complex

Ag(CN)2- <------>. Ag+ + 2CN-

At Equlibrium 1.0 X 10^-4M - x ------------ +x ------- 8 X 10^-4+2x

Kr = 1/Kf = 1.0 X 10^-21 = [Ag+][CN-]² / [Ag(CN)2-]

1.0 X 10^-21 = [x][8 X 10^-4+2x]² / [1.0 X 10^-4M - x ]

Kr is so small that the above equation can be simplified as

1.0 X 10^-21 = [x][8 X 10^-4 ]² / [1.0 X 10^-4M ]

x = 1.0 X 10^-21 X [1.0 X 10^-4M ] / [8 X 10^-4 ]² = 1.56 X 10^-11.

PPt of AgBr

Ag+ + Br- <----> AgBr

Qsp = [Ag+] [Br-] = 1.56 X 10^-11 X 2.05 X 10^-3 = 3.198 X 10^-14

Since Qsp < Ksp no precipitation will take place

1.0 x 10²¹ = [Ag(CN)₂^-] / [1.0 X 10^-4][1.0 X 10^-3M]²

[Ag(CN)₂^-] = 1.0 x 10²¹

queen_honey_blossom answered 7 months ago

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