Silver bromide AgBrs is insoluble and has a solubility produ

Silver bromide, AgBr(s), is insoluble and has a solubility product constant, Ksp = 5.4 x 10–13. Aqueous

silver ion forms a complex ion with CN– according to the following equation:

Ag+(aq) + 2 CN–(aq) --> Ag(CN)2–(aq) Kf = 1.2 x 1021

Given this information, determine the equilibrium constant for the reaction shown below.

AgBr(s) + 2 CN–(aq) --> Ag(CN)2–(aq) + Br–(aq)

Answer is 6.5 E8 - explanation gets points!

Solution

AgBr(s) ------ Ag+ + Br- ksp = 5.4* 10^-13

and

Ag+(aq) + 2 CN–(aq) --> Ag(CN)2–(aq) Kf = 1.2 x 10^21

adding above two equations will give the required equation, hance k value of required equation = k1*k2 = 5.4* 10^-13 *Kf = 1.2 x 10^21= 6.5 * 10^8