A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. 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))`

Options

A. (a)`4xx10^(-6)` M
B. (b)`2xx10^(-8)` M
C. (c )`2xx10^(-7)` M
D. (d)`8xx 10^(-15)` M

Answer

C

Explanation

`K_(sp) (AgCl)gt K_(sp)(Agl)`
Hence is `Agl` is precipitated before AgCl
`[Ag^(+)]` to percipitated `Cl^(-)=(1xx10^(-10))/0.05=2xx10^(-9) M`
`[I^(-)]` at this stage`=(K_(sp)(AgI))/([Ag^(+)])`
`=(4xx10^(-16))/(2xx10^(-9))=2xx10^(-7) M`