a) Sufficient sodium cyanide, NaCN, was added to 0.025 M silver nitrate, AgNO 3

Question

a)

Sufficient sodium cyanide, NaCN, was added to 0.025 M silver nitrate, AgNO3, to give a solution that was initially 0.150 M of a cyanide ion, CN-. What is the concentration of silver ion, Ag+, in this solution after Ag(CN)2- forms? The formation constant Kf for the complex ion Ag(CN)2- is 5.61E+18.

[Ag+] = _____M

b)

The formation constant Kf for the complex ion Zn(OH)42- is 2.8E+15. What is the concentration of zinc ion, Zn2+, in a solution that is initially 0.42M in Zn(OH)42-?

= ______M

c)

What is the molar solubility of NiS in 0.80 M NH3? The Ksp for NiS is 3.0E-19. The Kf for Ni(NH3)62+ is 5.6E+8.

= _____M

Explanation / Answer

Molarity . . . . .Ag+ + 2CN- ==> Ag(CN)2-
Initial . . . . 0.025 ..     0.15 . . . . . . . . .0
Change .     -0.025   ..-2(0.025)     0.025
Final                0        0.1                0.025

All of the Ag+ is used up, there is an excess of CN-, and 0.025 M Ag(CN)2- will be formed.

The Ag+ will be formed by dissociation of the complex

Dissociation constant of Ag(CN)2- = 1/Kf = 1 / (5.61 x 10^18) = 1.78 X 10^-19

Molarity . .. . . .Ag(CN)2- ==> Ag+ + 2CN-
Initial                0.025.                 0        0.1
Change                   x                      x          2x
At Equil.               0.025-x             x      0.1+2x

Kd = [Ag+][CN-]^2 / [Ag(CN)2-] = (x)(0.1+2x)^2 / (0.025-x) = 1.78 X 10^-19

With such a small Kd, the -x and +2x terms are negligible and can be dropped.

(x)(0.1) / (0.025) = 1.78 X 10^-19

x = 4.45 x 10^-20 M = [Ag+]

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