# The molar solubility of silver chromate at a certain temperature is 2 xx 10^(-4) "mol"cdot"dm"^(-3). What will it become in a solution containing "0.05 M" "AgNO"_3 at that same temperature?

Apr 8, 2016

Here's what I got.

#### Explanation:

Silver chromate, ${\text{Ag"_2"CrO}}_{4}$, is considered insoluble in aqueous solution, which means that in aqueous solution, an equilibrium will be established between the dissolved cations and anions and the undissolved solid.

${\text{Ag"_ color(red)(2)"CrO"_ (4(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CrO}}_{4 \left(a q\right)}^{2 -}$

By definition, the solubility product constant, ${K}_{s p}$, for this equilibrium will look like this:

${K}_{s p} = \left[{\text{Ag"^(+)]^color(red)(2) * ["CrO}}_{4}^{2 -}\right]$

Now, the problem provides you with the molar solubility of silver chromate at a given temperature, which essentially tells you how many moles of silver chromate can be dissolved in water before the resulting solution becomes saturated.

In this case, a molar solubility of $2 \cdot {10}^{- 4} {\text{mol dm}}^{- 3}$ tells you that you can dissolve $2 \cdot {10}^{- 4}$ moles of silver chromate per cubic decimeter ($\text{dm"^3 = "L}$) of solution at that given temperature.

To find the value of the solubility product constant, look at the mole ratios that exist between silver chromate and the dissolved ions, i.e. one mole of silver chromate produces two moles of silver cation and one mole of chromate anions.

If you can only dissolve $2 \cdot {10}^{- 4} \text{M}$ of silver chromate in aqueous solution, it follows that you will get

["Ag"^(+)] = color(red)(2) xx 2 * 10^(-4)"M" = 4 * 10^(-4)"M"

["CrO"_4^(2-)] = 1 xx 2 * 10^(-4)"M" = 2 * 10^(-4)"M"

This means that you have

${K}_{s p} = {\left(4 \cdot {10}^{- 4}\right)}^{\textcolor{red}{2}} \text{M"^color(red)(2) * 2 * 10^(-4)"M}$

${K}_{s p} = \textcolor{g r e e n}{\overline{\underline{|}} \stackrel{\text{ ")(" "3.2 * 10^(-11)"mol"^3"dm"^(-9)" }}{|}}$

To find the molar solubility of silver chromate in a solution that contains silver nitrate, ${\text{AgNO}}_{3}$, use the fact that this compound is soluble in aqueous solution

${\text{AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

Here one mole of silver nitrate produces one mole of silver cations, ${\text{Ag}}^{+}$. This means that the molarity of the silver cations before the silver chromate is added to the solution will be

["Ag"^(+)] = "0.05 M"

Keep in mind that the volume of the solution doesn't matter here because you're looking for concentration, not number of moles. In this regard, the volume could have been given as ${\text{100 dm}}^{3}$, as long as the molarity is said to be $\text{0.05 M}$, that's what you'll need to use.

Use an ICE table to find the new solubility

${\text{Ag"_ color(red)(2)"CrO"_ (4(s)) " "rightleftharpoons" " color(red)(2)"Ag"_ ((aq))^(+) " "+" " "CrO}}_{4 \left(a q\right)}^{2 -}$

color(purple)("I")color(white)(aaaaacolor(black)(-)aaaaaaaaaaaacolor(black)(0.05)aaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaacolor(black)(-)aaaaaaaaaaacolor(black)((+color(red)(2)s))aaaaaaaacolor(black)((+s))
color(purple)("E")color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)(color(red)(2)s + 0.05)aaaaaaaaaacolor(black)(s)

This time, you will have

${K}_{s p} = {\left(\textcolor{red}{2} s + 0.05\right)}^{\textcolor{red}{2}} \cdot s$

$3.2 \cdot {10}^{- 11} = \left(4 {s}^{2} + 0.2 s + 0.0025\right) \cdot s$

Rearrange to get

$4 {s}^{3} + 0.2 {s}^{2} + 0.0025 s - 3.2 \cdot {10}^{- 11} = 0$

This cubic equation will produce one real value

$s = 1.3 \cdot {10}^{- 8}$

Therefore, the molar solubility of silver chromate in a solution that contains $\text{0.05 M}$ of silver cations will be

$s = \textcolor{g r e e n}{\overline{\underline{|}} \stackrel{\text{ ")(" "1.3 * 10^(-8)"M"" }}{|}}$

I'll leave the answers rounded to two sig figs.

Notice that the solubility of silver chromate decreases significantly in the presence of the excess silver cations. We saw $2 \times {10}^{- 4}$ $\text{M} \to 1.3 \times {10}^{- 8}$ $\text{M}$.

This happens because the solubility equilibrium, which is governed by Le Chatelier's Principle, will lie further to the left in the presence of excess silver cations; hence, more ${\text{Ag"_2"CrO}}_{4}$ remains undissociated.

You'll see this referred to as the common-ion effect.