# 15.1 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.

## Check your learning

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?

## Answer:

[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 .

## Check your learning

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 .

## Answer:

1 $×$ 10 –10 M

#### Questions & Answers

what are oxidation numbers
Idowu Reply
pls what is electrolysis
Idowu Reply
Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.
AZEEZ
thanks
Idowu
what is the basicity of an atom
Eze Reply
basicity is the number of replaceable Hydrogen atoms in a Molecule. in H2SO4, the basicity is 2. in Hcl, the basicity is 1
Inemesit
how to solve oxidation number
Mr Reply
mention some examples of ester
Chinenye Reply
do you mean ether?
Megan
what do converging lines on a mass Spectra represent
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Bulus
oi
Amargo
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Bassidi
similarities between elements in the same group and period
legend Reply
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Nadeen
yvcrzezalakhhehuzhbshsunakakoaak
Nadeen
what is poh and ph
Amarachi Reply
please what is the chemical configuration of sodium
Sharon
2.8.1
david
1s²2s²2p⁶3s¹
Haile
2, 6, 2, 1
Salman
1s2, 2s2, 2px2, 2py2, 2pz2, 3s1
Justice
1s2,2s2,2py2,2
Maryify
1s2,2s2,2p6,
Francis
1s2,2s2,2px2,2py2,2pz2,3s1
Nnyila
what is criteria purity
Austin Reply
cathode is a negative ion why is it that u said is negative
Michael Reply
cathode is a negative electrode while cation is a positive ion. cation move towards cathode plate.
king
CH3COOH +NaOH ,complete the equation
david Reply
compare and contrast the electrical conductivity of HCl and CH3cooH
Sa Reply
The must be in dissolved in water (aqueous). Electrical conductivity is measured in Siemens (s). HCl (aq) has higher conductivity, as it fully ionises (small portion of CH3COOH (aq) ionises) when dissolved in water. Thus, more free ions to carry charge.
Abdelkarim
HCl being an strong acid will fully ionize in water thus producing more mobile ions for electrical conduction than the carboxylic acid
Valentine
differiante between a weak and a strong acid
david
how can I tell when an acid is weak or Strong
Amarachi
an aqueous solution of copper sulphate was electrolysed between graphite electrodes. state what was observed at the cathode
Bakanya Reply
write the equation for the reaction that took place at the anode
Bakanya
what is enthalpy of combustion
Bakanya
Enthalpy change of combustion: It is the enthalpy change when 1 mole of substance is combusted with excess oxygen under standard conditions. Elements are in their standard states. Conditions: pressure = 1 atm Temperature =25°C
Abdelkarim
Observation at Cathode: Cu metal deposit (pink/red solid).
Abdelkarim
Equation at Anode: (SO4)^2- + 4H^+ + 2e^- __> SO2 + 2H2O
Abdelkarim
Equation : CuSO4 -> Cu^2+ + SO4^2- equation at katode: 2Cu^2+ + 4e -> 2Cu equation at anode: 2H2O -> 4H+ + O2 +4e at the anode which reacts is water because SO4 ^ 2- cannot be electrolyzed in the anode
Niken

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