Question

For the reaction `Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19)` . Calculate `[Ag^(+)]` in solution which was originally `0*1` M in KCN and `0*03` M in `AgNO_(3)`.

Answer 1

Orignally on mixing KCN and `AgNO_(3)` , the reaction is
`{:(,2 KCN,+,AgNO_(3),to,"Ag"(CN)_(2)^(-),+,KNO_(3),+,K^(+)),("Intial amounts",0*1 M,,0*03 M,,0,,0,,0),("Amounts after reaction ",(0*1-0*06),,0,,0*03 M,,0*03M,,0*03),(,=0*04 M,,,,,,,,):}`
Thus , in the solution , now we have `[Ag (CN)_(2^(-)) ] = 0*3 M , [CN^(-)] = 0* 04 M .` Suppose x is the amount of `Ag(CN)_(2)^(-)` dissociated at equilibrium . Then
` {:(,AG(CN)_(2)^(-),hArr,AG^(+),+,2CN^(-)),("Intial amounts",0*03,,0,,0*04),("Amounts at eqm.",(0*03 -x),,x,+,0*04 + 2x):}`
` K_(c) = [ AG^(+) [CN^(-)]^(2))/([Ag(CN)_(2)^(-)])= (x (0*04 + 2x)^(2))/(0*03 - x) = 4 xx 10^(-19) ("Given")`
As `K_(c)` is very small , dissociation of `Ag(CN)_(2)^(-)` is very small i.e., x is very small . Hence , `0*04 + 2x approx and 0*3 - x approx 0*03 .`
` :. (x( 0*04)^(2))/(0*03 )= 4 xx 10^(-19) or x = 7*5 xx 10^(-18)`
Thus at equilibrium , `[Ag^(+)] = 7* 5 xx 10^(-18)M`