Question

Wanting to form nickel(II) cyanide, Ni(CN)2, a chemist adds hydrocyanic acid, HCN, to a solution containing Ni2+ ions hoping to precipitate Ni(CN)2. The important equlibria for this process are given below:
HCN(aq) <--> H+(aq) + CN-(aq) Ka=4.0x10^(-10)
Ni(CN)2(s) <--> Ni2+(aq) +2CN-(aq) Ksp= 3.0x1-^(-23)

a) what is the equilibrium constant, Knet, for the following reaction which describes the precipitation reaction the chemist wants to perform?
Ni2+(aq) + 2HCN(aq) <--> 2H+(aq) + Ni(CN)2(s)
Knet = ?

Answer 1

HCN(aq) <--> H+(aq) + CN-(aq) Ka=4.0x10^(-10) (Eq 1)
Ni(CN)2(s) <--> Ni2+(aq) +2CN-(aq) Ksp= 3.0x10^(-23) (Eq 2)

Multiply equation 1 by 2

2HCN(aq) <--> 2H+(aq) + 2CN-(aq) Ka=(4.0x10^(-10))^2 (Eq 3)

We have multiplied the equation by 2 , so Ka needs to be Ka^2
So we get

2HCN(aq) <--> 2H+(aq) + 2CN-(aq) Ka= 1.6x10^(-19) (Eq 3)

Now reverse the equation 2, we get

Ni2+(aq) +2CN-(aq) <--> Ni(CN)2(s) Ksp= 1/3.0x10^(-23) (Eq 4)

(Ksp for this equation is the reciprocal for the Ksp of equation 2)

So equation 4 becomes

Ni2+(aq) +2CN-(aq) <--> Ni(CN)2(s) Ksp= 3.3x10^(22) (Eq 4)

Addiing equation 3 & 4, we get the desired reaction

Ni2+(aq) + 2HCN(aq) <--> 2H+(aq) + Ni(CN)2(s) (Eq 5)

Knet = product of equilibrium constants of equation 3 & 4

So

Knet = 1.6x10^(-19) x 3.3x10^(22)

K net = 5333

So for

Ni2+(aq) + 2HCN(aq) <--> 2H+(aq) + Ni(CN)2(s)   K net = 5333 (Eq 5)