Question

A solution is a mixture of $0.06$M $KCl$ and $0.06$M $KI$. $AgN{O}_3$ solution is being added drop by drop till $AgCl$ starts precipitating $(K_{sp}AgCl=1\times {10}^{-10}$ and $K_{sp}AgI=4\times {10}^{-16})$. The concentration of Iodide ion at this stage will be nearly equal to:

• A. $4.0\times {10}^{-5}$M
• B. $2.4\times {10}^{-7}$M
• C. $2.0\times {10}^{-8}$M
• D. $4\times {10}^{-8}$M

Answer 1

The correct option is B $2.4\times {10}^{-7}$M

After you mix the two solutions, the resulting solution will contain two anions that can combine with the silver(I) cations to form an insoluble precipitate, the chloride anions and the iodide anions, respectively.

For silver chloride, the solubility equilibrium looks like this-

$AgCl⟶A{g}^{+}+C{l}^{-}$

Solubility product for $AgCl$

$K_{sp}AgCl=\left[Ag{]}^{+}\right[Cl{]}^{-}$

Now there is two conditions that $A{g}^{+}$ ions can combine with $C{l}^{-}$ or $I^{-}$ ions,

In your case, you know that the solution contains $C{l}^{-}=.06M$

This means that the minimum concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to-

$[Ag{]}_{min1}^{+}={\frac{\left[K_{sp}AgCl\right]}{\left[Cl\right]}}^{-}$

$[Ag{]}_{min1}^{+}=\frac{{10}^{-10}}{.06}$

Now for silver iodide solubility equilibrium will look like this-

$AgI\to A{g}^{+}+I^{-}$

This means that the minimum concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to-

$[Ag{]}_{min2}^{+}=\frac{\left[K_{sp}AgI\right]}{\left[I^{-}\right]}$

$[Ag{]}_{min2}^{+}=\frac{4\times {10}^{-16}}{.06}$

Notice that we have,

$[Ag{]}_{min2}^{+}<<<[Ag{]}_{min1}^{+}$

This tells you that as you start to add silver(I) cations to the solution, the silver iodide will precipitate first, consuming some of the iodide ions present in the solution in the process.

you need a concentration of silver(I) cations equal to start precipitating $AgCl$

$[Ag{]}_{min.1}^{+}=2\times {10}^{-9}$

present in the solution. This means that at that point, the corresponding concentration of iodide anions must be at a maximum--which corresponds to the solubility equilibrium condition value of

$[I{]}^{-}={\frac{K_{sp}AgI}{\left[Ag\right]}}_{min1}^{+}=2.4\times {10}^{-7}M$