Question

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaICuIAgICalculate [Ag+]when CuI just begins to precipitate.×10MEnter your answer in scientific notation.What percent of Ag+ remains in solution at this point?

Answer 1

AgI should precipitate first.

The concentration of
`Ag^+` when CuI just begins to precipitate is,
`6.64* 10^(-7)M`

Percent of
`Ag^+` remains is, 0.0076 %

Explanation :

`K_(sp)` for CuI is
`1* 10^(-12)`

`K_(sp)` for AgI is
`8.3* 10^(-17)`

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

`CuI\rightleftharpoons Cu^++I^-`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Cu^+][I^-]`

`1* 10^(-12)=0.0079* [I^-]`

`[I^-]=1.25* 10^(-10)M`

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

`AgI\rightleftharpoons Ag^++I^-`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Ag^+][I^-]`

`8.3* 10^(-17)=[Ag^+]* 1.25* 10^(-10)M`

`[Ag^+]=6.64* 10^(-7)M`

Now we have to calculate the percent of
`Ag^+` remains in solution at this point.

Percent of
`Ag^+` remains =
`(6.64* 10^(-7))/(0.0087)* 100`

Percent of
`Ag^+` remains = 0.0076 %