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VMariaS [17]
3 years ago
9

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?NaI

CuIAgICalculate [Ag+]when CuI just begins to precipitate.×10MEnter your answer in scientific notation.What percent of Ag+ remains in solution at this point?
Chemistry
1 answer:
NeTakaya3 years ago
8 0

Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 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\times 10^{-12}=0.0079\times [I^-]

[I^-]=1.25\times 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\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

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

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

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