Calculate the solubility of AgCN in a buffer solution of pH 3.00 assuming no complex formation, `K_(a) (HCN)=6.2 xx 10^(-10) and K_(sp) (AgCN)=2.2 xx 10^(-16)`.

Explanation

Let the solubility of AgCN be x moles/litre. On dissolution silver remains as `Ag^(+)" but "CN^(-)` ions are converted mostly to HCN due to the fixed acidity of the buffer.
`K_(a)=([H^(+)] [CN^(-)])/([HCN])`
`or ([HCN])/([CN^(-)])=([H^+])/(K_(a))=(1 xx 10^(-3))/(6.2 xx 10^(-10))=1.6 xx 10^(6)........(1)`
As each `CN^(-)` ion hydrolysis to yeild one HCN
`x=[Ag^(+)] =[CN^(-)]+[HCN]`
from the ratio `[HCN]//[CN^(-)], we see "[CN^(-)] lt lt [HCN]`
`x=[Ag^(+)]=[HCN]`
Thus from the equation (1) we have
`[CN^(-)]=(x)/(1.6 xx 10^(6))`
Now, `K_(sp)=[Ag^(+)] [CN^(-)]`
`2.2 xx 10^(-16) =x(x)/(1.6 xx 10^(-6))`
`x=1.9 xx 10^(-5)M`