Question

Calculate the [Ag⁺] in a solution prepared by dissolving 1.00g of AgNO₃ and 10.0g KCN in enough water to make a 1.00 L of solution Kf[Ag(CN)₂]^1-=1.0 x 10²¹

Answer: [Ag⁺] = 2.9 x 10^-22M

Please show the steps required to get to the answer.

Answer 1

AgNO3 moles = mass of AgNO3 / molar mass of AgNO3= 1/ 169.87 = 0.005887

KCN moles = 10 / 65.12 = 0.15356

Ag+ (aq) + 2CN-(aq) <----> Ag(CN)2 (s)

as per reaction 1AgNo3 reacts with 2KCN t form complex.

moles of AgNO3 =0.005887 moles of kCN erequired = 2 x 0.005887 = 0.011774 but we had 0.15356 moles KCN . Hene KCn is in excess and AgNO3 is limiting reagent.

Ag(CN)2 moles formed = AgNO3 moels reacted = 0.005887

total vol = 1L

now we have Ag(CN)2(s) <---> Ag+ (aq) + 2CN- (aq)

K = 1/Kf = ( 1/10^ 21) = 10^ -21

Hence K = [Ag+] [CN-]^2 / [Ag(CN)2]-1= 10^ -21    

( we had [CN-] = ( 0.15356 - 2(0.005887) = 0.141786 i.e CN- left after reaction with Ag+)

1Now 0^ -21 = [Ag+] ( 0.141786)^2 / ( 0.005887)

[Ag+] = 0.29 x 10^ -21 M = 2.9 x 10^ -22 M