Question

The solubility product of CuI is 1.0×10–12. The formation constant for the reaction of CuI with I– to give CuI2– is 7.9×10–4. Calculate the molar solubility of CuI in a 1.0×10–4M solution of KI.

Answer 1

Explanation:

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Answer 2

The molar solubility  $CuI$ will be 1 x 10 to

Given:

The solubility product $(K_{sp})$ = $1$ × $10^{-12}$

The formation constant for the reaction $(K_{f})$ = $7.9$ × $10^{-4}$

The concentration of KI = $1$ × $10^{-4}$

To Find:

The molar solubility of  $CuI$ in a $1$ × $10^{-4}$ solution of KI.

Solution:

The answer to this question can be found very easily as given below,

Here,

This is the reaction between $Cu^{2+}$ and $I^{-}$ to get $CuI$

So, the equation of this reaction will be,

$CuI^{-} _{2}$ ⇄ $Cu^{2+} + 2I^{-}$

While we consider the Initial, Change, and Equilibrium concept (ICE)

The initial concentration of $Cu^{2+}$ will be zero

The initial concentration of $2I^{-}$ will be $1$ × $10^{-4}$

Now consider,

The change in concentration for $Cu^{2+}$ as $x$

The change in concentration for $2I^{-}$ as $2x$

Then at equilibrium,

The concentration of $Cu^{2+}$ will be $x$

The concentration of $2I^{-}$ will be $1$ × $10^{-4} + 2x$

We know that,

The equilibrium constant for the dissolution of a solid substance into an aqueous solution is the solubility product constant.

That is,

$K_{sp} = [Cu^{2+}] [I^{-}]^{2}$

By substituting the values,

$1$ × $10^{-12}$ = $[x]$ $[10^{-4} + 2x]^{2}$

Assume that $[10^{-4} + 2x]$ ≈ $10^{-4}$ (It is insignificant)

Therefore,

$1$ × $10^{-12}$ = $x$ × $(10^{-4})^{2}$

$x$ = $\frac{10^{-12} }{10^{-16} }$

$x$ = $10^{4}$

Hence, the molar solubility of $CuI$ will be $10^{4}$

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