# 6.5 Precipitation and dissolution  (Page 8/17)

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## Precipitation of silver halides

A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO 3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

## Solution

The two equilibria involved are:

$\text{AgCl}\left(s\right)⇌{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{−}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
$\text{AgI}\left(s\right)⇌{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{−}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-16}$

If the solution contained about equal concentrations of Cl and I , then the silver salt with the smallest K sp (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag + ] at which AgCl begins to precipitate and the [Ag + ] at which AgI begins to precipitate. The salt that forms at the lower [Ag + ] precipitates first.

For AgI: AgI precipitates when Q equals K sp for AgI (1.5 $×$ 10 –16 ). When [I ] = 0.0010 M :

$Q=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{I}}^{\text{−}}\right]=\left[{\text{Ag}}^{+}\right]\text{(0.0010)}=\text{1.5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-16}$
$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{\text{1.8}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}}{0.10}\phantom{\rule{0.2em}{0ex}}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$

AgI begins to precipitate when [Ag + ] is 1.5 $×$ 10 –13 M .

For AgCl: AgCl precipitates when Q equals K sp for AgCl (1.6 $×$ 10 –10 ). When [Cl ] = 0.10 M :

${Q}_{\text{sp}}=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{Cl}}^{\text{−}}\right]=\left[{\text{Ag}}^{\text{+}}\right]\text{(0.10)}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}}{0.10}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$

AgCl begins to precipitate when [Ag + ] is 1.6 $×$ 10 –9 M .

AgI begins to precipitate at a lower [Ag + ] than AgCl, so AgI begins to precipitate first.

If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag + ] would precipitation begin, and what would be the formula of the precipitate?

[Ag + ] = 1.0 $×$ 10 –11 M ; AgBr precipitates first

## Common ion effect

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH 3 CO 2 , is added. We can explain this effect using Le Châtelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ to compensate for the increased acetate ion concentration. This increases the concentration of CH 3 CO 2 H:

${\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{H}}_{2}\text{O}⇌{\text{H}}_{3}{\text{O}}^{\text{+}}+{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}$

Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect    .

The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:

$\text{AgI}\left(s\right)⇌{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{−}}\left(aq\right)$

If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Châtelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.

## Common ion effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010- M solution of cadmium bromide (CdBr 2 ). The K sp of CdS is 1.0 $×$ 10 –28 .

## Solution

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr 2 concentration as a contributor of cadmium ions:

$\text{CdS}\left(s\right)⇌{\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2−}}\left(aq\right)$
${K}_{\text{sp}}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{\text{2−}}\right]=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$
$\left(0.010+x\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$
${x}^{2}+\text{0.010}x-\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}=0$

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the K sp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our K sp expression, we would now get:

${K}_{\text{sp}}=\left[{\text{Cd}}^{\text{2+}}\right]{\left[\text{S}}^{\text{2−}}\right]=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$
$\left(0.010\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$
$x=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}$

Therefore, the molar solubility of CdS in this solution is 1.0 $×$ 10 –26 M .

Calculate the molar solubility of aluminum hydroxide, Al(OH) 3 , in a 0.015- M solution of aluminum nitrate, Al(NO 3 ) 3 . The K sp of Al(OH) 3 is 2 $×$ 10 –32 .

1 $×$ 10 –10 M

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