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=????? b) can the chemist readily form Nickel (II) Cyanide in this manner? c) The chemist runs the reaction and allows the solution to come to equilibrium before diluting the solution by a factor of 10. Calculate Q upon dilu Also, will he get more or less Ni(CN)2 by diluting the solution in this fashion? Can you please explain how you came up with the answer. I think I have parts a and b, but am having trouble with part c.

Answer 1

c) Ni2+(aq) + 2HCN(aq) <--> 2H+(aq) + Ni(CN)2(s) Knet=?????

Ka=[H+][CN-]/[HCN]

Ksp=[Ni2+][CN-]^2

Knet=[H+]^2/[Ni2+][HCN]^2

Ka^2/Ksp=[H+]^2[CN-]^2/[HCN]^2*(1/[Ni2+][CN-]^2) =[H+]^2/[Ni2+][HCN]^2=knet

knet= Ka^2/Ksp =(4.0*10^-10)^2/3.0*10^-23=5.3*10^-3

now you have, Ni2+(aq) + 2HCN(aq) <--> 2H+(aq) + Ni(CN)2(s) knet=5.3*10^-3

Q=reaction quotient=[H+]^2/[Ni2+][HCN]^2

When diluted by a factor of 10,

[H+]=1/10[H+]

[Ni2+]=1/10[Ni2+]

[HCN]=1/10[HCN]

Q’=[H+]^2/[Ni2+][HCN]^2 *[(1/10)^2/(1/10)*(1/10)^2=

     =[H+]^2/[Ni2+][HCN]^2 *[1/(1/10)]

   ==[H+]^2/[Ni2+][HCN]^2 *10=10Q(becomes 10 times)

As Q’=10 Q so the reaction equilibrium moves in forward direction , giving more Ni(CN)2