Question

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.160-mole quantity of M(NO3)2 is added to a liter of 1.070 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Answer 1

If you put 0.160 moles of M(NO3) in 1 L of solution, then [M+] = 0.160 M, not considering its reaction with CN- to form the [M(CN)4]2– complex ion.
As the Kf value is very large, we can say that the reaction is quantitative, being the CN- concentration in significant excess. Therefore we get 0.160 M [M(CN)4]2– , together with a very small concentration of M+ ion left in solution.

M(+) + 4CN(-) <-------> [M(CN)4]2(-)
0.160 . . .1.07 . . . . . . . . .0
.0.160-x . .4*(0.160-x) . . . .0.160

Kf = [[M(CN)4]2(-)]/ [M+][CN-]^4 = 0.160 / (x*(0.640 + 4x)^4) = 7.70 * 10^116

We know that x will be very small, so 4x will be very small compared to 0.64. Then we can delete the 4x term to simplify the equation.

0.160 / (x*(0.64)^4) = 7.7 * 10^16

x = 0.160/(0.64^4*7.7 * 10^16) = [M+] = 1.238*10^-17 M