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Law Incorporation [45]
2 years ago
7

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

t?AgBrCuBrWhat percent of Ag remains in solution at the point when CuBr just begins to precipitate?
Chemistry
1 answer:
saul85 [17]2 years ago
3 0

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

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

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

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

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

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