Question

The Ksp of AgCl is 1.8x10^-10 and the Ksp of AgI is 8.3x10^-17. A solution is .100M in I- and Cl-. When a silver nitrate solution is slowly added to this mixture, what is the molarity of iodide ions when AgCl just starts to precipitate?

A.) 1.0x10^-5M
B.) 9.1x10^-9M
C.) 8.3x10^-7M
D.) 4.6x10^-8M

Answer 1

`[I^-]=4.6*10^(-8)M`

Step-by-step explanation:

The expression of the Ksp:

`Ksp_(AgCl)=[Ag^(+)][Cl^-]`

`Ksp_(AgI)=[Ag^(+)][I^-]`

When the product of the concentrations of both ions equals the Ksp, the salt starts to precipitate.

For the AgCl:

`1.8*10^(-10)M^(2)=[Ag^(+)]*0.1M`

`[Ag^(+)]=1.8*10^(-9)M`

Initially the concentration of I- was 0.1 M, due to the lower Ksp than the AgCl's, the AgI will precipite before. So, when AgCl starts to precipitate the concentration of I- will be in equilibrium, following the Ksp equation.

`8.3*10^(-17)M^(2)=1.8*10^(-9)M*[I^-]`

`[I^-]=4.6*10^(-8)M`