Common Ion Effect Revision Notes for HSC SSCE Chemistry

Common Ion Effect

Introduction to the common ion effect

When studying solution equilibria, you may have learned about situations where a single ionic salt dissolves in water. The common ion effect introduces an important variation on this concept. This effect occurs when you add a substance that shares an ion with a sparingly soluble salt already present in solution.

For example, imagine you have a test tube containing calcium sulfate ( $\text{CaSO}_4$ ), which is sparingly soluble in water. If you then add sodium sulfate ( $\text{Na}_2\text{SO}_4$ ) to this test tube, both compounds share the sulfate ion ( $\text{SO}_4^{2-}$ ). This shared ion is called a common ion, and its presence affects the solubility of the original salt.

chatImportant

Key Principle: Adding a common ion to a saturated solution decreases the solubility of the salt.

Understanding the effect using Le Chatelier's principle

To understand why the common ion effect occurs, we can apply Le Chatelier's principle to a solubility equilibrium.

Consider calcium sulfate dissolving in water:

$\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

This equilibrium exists when calcium sulfate is in a saturated solution.

What happens when we add sodium sulfate?

Sodium sulfate ( $\text{Na}_2\text{SO}_4$ ) is a soluble salt. When added to water, it completely dissociates:

$\text{Na}_2\text{SO}_4(s) \rightarrow 2\text{Na}^{+}(aq) + \text{SO}_4^{2-}(aq)$

When you add sodium sulfate to the test tube containing calcium sulfate, you are increasing the concentration of sulfate ions ( $\text{SO}_4^{2-}$ ) in the solution.

Le Chatelier's principle in action

According to Le Chatelier's principle, when you increase the concentration of one component in an equilibrium system, the system responds by shifting to reduce that concentration. In this case:

The increased concentration of $\text{SO}_4^{2-}$ causes the equilibrium to shift backwards (to the left)
This shift favours the reverse reaction
More solid calcium sulfate precipitates out of solution
The concentration of $\text{Ca}^{2+}$ ions decreases
The overall result is that calcium sulfate becomes less soluble

infoNote

This qualitative explanation shows why the common ion effect occurs: the system responds to the added ion by shifting the equilibrium to reduce its concentration, which means more solid precipitates and overall solubility decreases.

Quantitative analysis of the common ion effect

Now let's examine the common ion effect using calculations. This will show you exactly how much the solubility changes.

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Worked Example: Calcium Sulfate in Water versus in Sodium Sulfate Solution

The solubility product constant ( $K_{sp}$ ) for calcium sulfate at $25°\text{C}$ is $4.93 \times 10^{-5}$ .

Part A: Solubility in pure water

For the equilibrium:

$\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

The equilibrium expression is:

$K_{sp} = [\text{Ca}^{2+}] \times [\text{SO}_4^{2-}]$

In pure water, when calcium sulfate dissolves, it produces equal concentrations of calcium and sulfate ions. If we call this concentration $s$ (representing solubility):

$[\text{Ca}^{2+}] = [\text{SO}_4^{2-}] = s$

Substituting into the $K_{sp}$ expression:

$K_{sp} = 4.93 \times 10^{-5} = s^2$

Solving for $s$ :

$s = \sqrt{4.93 \times 10^{-5}} = 7.02 \times 10^{-3} \text{ mol L}^{-1}$

Result: In pure water, the calcium ion concentration is $7.02 \times 10^{-3} \text{ mol L}^{-1}$ .

Part B: Solubility in sodium sulfate solution

Now let's calculate what happens when calcium sulfate dissolves in a solution that already contains $0.100 \text{ mol L}^{-1}$ sodium sulfate.

Since sodium sulfate completely dissociates and has a 1:1 ratio with sulfate ions:

$[\text{SO}_4^{2-}] = [\text{Na}_2\text{SO}_4] = 0.100 \text{ mol L}^{-1}$

When calcium sulfate dissolves in this solution:

$[\text{Ca}^{2+}] = s$ (from dissociation of $\text{CaSO}_4$ )
$[\text{SO}_4^{2-}] = (0.100 + s)$ (from both sodium sulfate and calcium sulfate)

However, because sodium sulfate completely dissociates and calcium sulfate only sparingly dissociates, the value of $s$ is much smaller than $0.100 \text{ mol L}^{-1}$ . This means:

$s \ll 0.100 \text{ mol L}^{-1}$

Therefore, we can make a simplifying assumption:

$(0.100 + s) \approx 0.100$

Substituting into the equilibrium expression:

$K_{sp} = 4.93 \times 10^{-5} = s \times 0.100$

Solving for $s$ :

$s = \frac{4.93 \times 10^{-5}}{0.100} = 4.93 \times 10^{-4} \text{ mol L}^{-1}$

Result: In $0.100 \text{ mol L}^{-1}$ sodium sulfate solution, the calcium ion concentration is $4.93 \times 10^{-4} \text{ mol L}^{-1}$ .

Comparing the results

Notice that $s = 4.93 \times 10^{-4}$ is indeed much smaller than $0.100$ , which validates our simplifying assumption.

chatImportant

Key Observation: The calcium ion concentration decreased from $7.02 \times 10^{-3}$ to $4.93 \times 10^{-4}$ mol L $^{-1}$ when the common ion was present. This confirms that calcium sulfate is less soluble when a common ion is present, supporting our qualitative prediction using Le Chatelier's principle.

Worked example: silver iodide solubility

Let's examine another detailed example comparing the solubility of silver iodide in water versus in a potassium iodide solution.

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Worked Example: Silver Iodide Solubility with Common Ion Effect

Problem Statement

Compare both qualitatively and quantitatively the solubility of silver iodide in water with its solubility in $0.250 \text{ mol L}^{-1}$ potassium iodide ( $\text{KI}$ ). The $K_{sp}$ for silver iodide is $8.52 \times 10^{-17}$ at $25°\text{C}$ .

Solution

Step 1: Write the balanced equation

$\text{AgI}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{I}^{-}(aq)$

Step 2: Qualitative analysis using Le Chatelier's principle

When potassium iodide is added to a solution containing silver iodide, the concentration of iodide ions ( $\text{I}^{-}$ ) increases. According to Le Chatelier's principle, the equilibrium system will respond by shifting backwards (favouring the reverse reaction). This increases precipitation of silver iodide, meaning the solubility of silver iodide decreases.

Step 3: Write the equilibrium expression

$K_{sp} = [\text{Ag}^{+}] \times [\text{I}^{-}]$

Step 4: Calculate solubility in water

In pure water:

$[\text{Ag}^{+}] = [\text{I}^{-}] = s$

Therefore:

$K_{sp} = 8.52 \times 10^{-17} = s^2$

$s = \sqrt{8.52 \times 10^{-17}} = 9.23 \times 10^{-9} \text{ mol L}^{-1}$

Solubility of $\text{Ag}^{+}$ in water is $9.23 \times 10^{-9}$ mol L $^{-1}$ .

Step 5: Calculate solubility in potassium iodide solution

Since potassium iodide completely dissociates with a 1:1 ratio:

$[\text{I}^{-}] = [\text{KI}] = 0.250 \text{ mol L}^{-1}$

When silver iodide dissolves:

$[\text{Ag}^{+}] = s$ (from dissociation of $\text{AgI}$ )
$[\text{I}^{-}] = (0.250 + s)$ (from both $\text{KI}$ and $\text{AgI}$ )

Since $s \ll 0.250$ , we can approximate:

$(0.250 + s) \approx 0.250$

Substituting into the equilibrium expression:

$K_{sp} = 8.52 \times 10^{-17} = s \times 0.250$

$s = \frac{8.52 \times 10^{-17}}{0.250} = 3.41 \times 10^{-16} \text{ mol L}^{-1}$

Solubility of $\text{Ag}^{+}$ in $0.250$ mol L $^{-1}$ potassium iodide is $3.41 \times 10^{-16}$ mol L $^{-1}$ .

Step 6: Compare the values

To see how much the solubility decreased:

$\frac{9.23 \times 10^{-9}}{3.41 \times 10^{-16}} = 27 \times 10^{6} = 27,000,000$

Conclusion: The solubility of silver iodide is approximately 27 million times less in $0.250$ mol L $^{-1}$ potassium iodide compared to its solubility in water.

This dramatic decrease clearly demonstrates the powerful effect of the common ion on solubility.

Practical investigation: demonstrating the common ion effect

Investigation 4.5: Teacher demonstration

Aim

To observe the common ion effect in action using a saturated sodium chloride solution.

Materials required

$6 \text{ mL}$ saturated $\text{NaCl}$ solution
Dropper bottle containing concentrated $\text{HCl}$
Dropper bottle containing $10 \text{ mol L}^{-1}$ $\text{NaOH}$
3 test tubes
Test tube rack
3 small labels
Measuring cylinder ( $5 \text{ mL}$ ) or disposable droppers
Safety glasses

Risk assessment

chatImportant

Key Safety Considerations:

Concentrated hydrochloric acid is highly corrosive - use in a fume cupboard or well-ventilated area
$10 \text{ mol L}^{-1}$ sodium hydroxide is very corrosive - wear safety glasses
Always wash hands immediately after handling chemicals

Method

Label the three test tubes A, B and C
Add $2 \text{ mL}$ of saturated sodium chloride solution to each test tube
To test tube B, add 10 drops of concentrated hydrochloric acid. Record your observations
To test tube C, add 10 drops of $10 \text{ mol L}^{-1}$ sodium hydroxide. Record your observations
Compare the three test tubes

What to observe

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Understanding the Setup:

Test tube A serves as the control (no additions made)
Test tube B receives $\text{HCl}$ , which provides additional chloride ions ( $\text{Cl}^{-}$ ) - the common ion
Test tube C receives $\text{NaOH}$ , which provides additional sodium ions ( $\text{Na}^{+}$ ) - the common ion

When a common ion is added to either test tube B or C, you should observe increased precipitation of sodium chloride crystals due to the common ion effect.

Results and conclusion

Record your observations in a carefully designed table. Consider these questions:

What was the purpose of test tube A? (It serves as a control for comparison)
How does this investigation illustrate the common ion effect? (Adding a common ion causes more precipitation)

Investigation 4.6: Determining $K_{sp}$ for lead(II) iodide

Introduction

This investigation allows you to determine the solubility product constant indirectly by preparing mixtures with different concentrations of lead ions ( $\text{Pb}^{2+}$ ) and iodide ions ( $\text{I}^{-}$ ).

The equilibrium is:

$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^{-}(aq)$

The ionic product is:

$Q_{sp} = [\text{Pb}^{2+}] \times [\text{I}^{-}]^2$

infoNote

In some test tubes, $Q_{sp}$ will exceed $K_{sp}$ and a precipitate will form. In others, $Q_{sp}$ will be less than $K_{sp}$ and no precipitate will form. By finding the boundary between these two cases, you can determine the range for $K_{sp}$ .

Aim

To determine the $K_{sp}$ for lead(II) iodide.

Materials required

Burette containing $1.00 \times 10^{-2} \text{ mol L}^{-1}$ $\text{Pb(NO}_3\text{)}_2$
Burette containing $1.00 \times 10^{-2} \text{ mol L}^{-1}$ $\text{KI}$
Burette containing deionised water
9 large test tubes
Test tube rack
9 small labels
Heavy metal waste bottle
Retort stands and burette clamps
Safety glasses

Risk assessment

chatImportant

Key Safety Considerations:

Lead salts are highly toxic - use burettes for dispensing
Always wear safety glasses
Wash hands immediately after use
Dispose of all solutions in the heavy metal waste bottle provided

Method

Label each test tube from 1 to 9
Prepare each mixture according to the volumes specified in the table below
Always add the water to the test tube first (this prevents local supersaturation)
Leave each mixture for at least 30 minutes before checking for precipitate formation

Analysis of results

Step 1: Calculate ion concentrations

For each test tube, calculate the concentration of each ion using:

$[\text{ion}]_{\text{mixture}} = [\text{ion}]_{\text{stock solution}} \times \frac{\text{Volume (mL) of ion added}}{\text{Total volume (mL) of mixture}}$

Step 2: Calculate $Q_{sp}$ values

Use the calculated ion concentrations to determine $Q_{sp}$ for each mixture:

$Q_{sp} = [\text{Pb}^{2+}] \times [\text{I}^{-}]^2$

Step 3: Determine the $K_{sp}$ range

Identify the smallest $Q_{sp}$ value where a precipitate formed
Identify the largest $Q_{sp}$ value where no precipitate formed
The $K_{sp}$ lies between these two values

Conclusion

Compare your experimental range for $K_{sp}$ with the literature value provided on the NESA data sheet. Discuss possible reasons for any differences, such as:

Temperature variations
Experimental uncertainties in volume measurements
Time allowed for equilibrium to establish
Purity of reagents

Summary of key concepts

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Essential Concepts of the Common Ion Effect:

Definition: The common ion effect occurs when adding a common ion to a saturated solution decreases the solubility of the salt.

Mechanism: This effect can be explained using Le Chatelier's principle. When you increase the concentration of one ion in an equilibrium system, the system shifts to reduce that concentration, favouring precipitation.

Quantitative impact: Calculations show that the decrease in solubility can be dramatic. For example, silver iodide is 27 million times less soluble in potassium iodide solution compared to pure water.

Practical applications: The common ion effect is important in many areas of chemistry, including precipitation reactions, buffer solutions, and analytical chemistry.

Remember!

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Key Points to Remember:

Adding a common ion always decreases solubility - when you add a substance containing an ion that's already present in a saturated solution, the sparingly soluble salt becomes even less soluble
Le Chatelier's principle explains why - increasing the concentration of one component causes the equilibrium to shift backwards, favouring precipitation and reducing solubility
The effect is quantifiable - you can calculate exactly how much the solubility changes using the solubility product constant ( $K_{sp}$ ) and the concentration of the added common ion
Use the simplifying assumption - when the common ion comes from a completely dissociating salt, its concentration is much larger than the contribution from the sparingly soluble salt, so you can approximate $(c + s) \approx c$
Safety is essential - when working with chemicals in investigations, always follow proper safety procedures, wear appropriate protective equipment, and dispose of waste correctly

Common Ion Effect (HSC SSCE Chemistry): Revision Notes