Question

A solution is a mixture of 0.05 M NaCl and 0.05 M AgI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to:
`(K_(sp)AgCl=1xx10^(-10)M^(2), K_(sp)AgI=4xx10^(-16)M^(2))`
A. `4xx10^(-6)M`
B. `2xx10^(-8)M`
C. `2xx10^(-7)M`
D. `8xx10^(-15)M`

Answer 1

Correct Answer - C
`[Ag^(+)]=(K_(SP)AgCI)/([CI^(-)])` For AgCI precipitation
`= (10^(-10))/(0.05)=2xx10^(-9)`
`:. [Ag^(+)]=(K_(SP)AgI)/([I^(-)])` For AgI precipitation
`= (4xx10^(-16))/(0.05)=8xx10^(-15)`
Thus, AgI will precipitate first. AgCI will precipitate only, when
`[Ag^(+)]=2xx10^(-9)`, Thus
`[I^(-)]_(l eft)=(4xx10^(-16))/(2xx10^(-9))=2xx10^(-7)M`