Question

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

ksp nickle (II) cyanide --> 3.0 × 10^-23

Answer 1

Answer : The maximum amount of nickel(II) cyanide is
`5.84* 10^(-12)M`

Explanation :

The solubility equilibrium reaction will be:

`Ni(CN)_2\rightleftharpoons Ni^(2+)+2CN^-`

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

`K_(sp)=[Ni^(2+)][CN^-]^2`

Now put all the given values in this expression, we get:

`3.0* 10^(-23)=(0.220+s)* (2s)^2`

`s=5.84* 10^(-12)M`

Therefore, the maximum amount of nickel(II) cyanide is
`5.84* 10^(-12)M`