# A solution is a mixture of "0.05M NaCl" and "0.05M NaI".The concentration of iodide ion in the solution when "AgCl" just starts precipitating is equal to?

## ${K}_{\text{sp AgCl}} = 1 \cdot {10}^{- 10}$ ${\text{M}}^{2}$ ${K}_{\text{sp AgI}} = 4 \cdot {10}^{- 16}$ ${\text{M}}^{2}$

Feb 10, 2018

$\left[{\text{I}}^{-}\right] = 2 \cdot {10}^{- 7}$ $\text{M}$

#### Explanation:

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

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

By definition, the solubility product constant for this equilibrium is equal to

${K}_{{\text{sp AgCl") = ["Ag"^(+)] * ["Cl}}^{-}}$

In your case, you know that the solution contains

["Cl"^(-)] = "0.05 M"

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

["Ag"^(+)]_ "min 1" = K_("sp AgCl")/(["Cl"^(-)])

["Ag"^(+)]_ "min 1" = (1 * 10^(-10) "M"^color(red)(cancel(color(black)(2))))/(0.05color(red)(cancel(color(black)("M")))) = 2 * 10^(-9) quad "M"

$\textcolor{w h i t e}{a}$
Now, for silver iodide, the solubility equilibrium looks like this

${\text{AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I}}_{\left(a q\right)}^{-}$

This time, the solubility product constant is equal to

${K}_{{\text{sp AgI") = ["Ag"^(+)] * ["I}}^{-}}$

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

["Ag"^(+)]_ "min 2" = K_("sp AgI")/(["I"^(-)])

["Ag"^(+)]_"min 2" = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(0.05 color(red)(cancel(color(black)("M")))) = 8 * 10^(-15) quad "M"

Notice that you have

["Ag"^(+)]_ "min 2" " << " ["Ag"^(+)]_"min 1"

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.

This tells you that when silver chloride precipitate, the solution will not contain

["I"^(-)] = "0.05 M"

because some of the iodide ions already precipitated as silver iodide. In order for the silver(I) cations to precipitate the silver chloride when

["Cl"^(-)] = "0.05 M"

you need a concentration of silver(I) cations equal to

["Ag"^(+)]_ "min1" = 2 * 10^(-9) quad "M"

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

$\left[\text{I"^(-)] = K_("sp AgI")/(["Ag"^(+)]_ "min 1}\right)$

["I"^(-)] = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(2 * 10^(-9) color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(2 * 10^(-7) quad "M")))

The answer is rounded to one significant figure.