Why Percentage Ionisation Increases When a Weak Acid Is Diluted

Take 25.0 mL of 0.100 mol L\(^{-1}\) ethanoic acid and dilute it to 250.0 mL. The solution becomes less acidic, but a greater percentage of the acid molecules are now ionised.

Before reading on, make a prediction: does the concentration of hydronium ions, \([\ce{H3O+}]\), increase or decrease?

It decreases. The percentage ionisation increases because dilution shifts the acid equilibrium towards the formation of ions. The important distinction is that a greater fraction of acid molecules ionises, even though the number of hydronium ions per litre becomes smaller.

Let’s work through why this happens, how to calculate the change, and where the usual approximation stops being reliable.

01Why does dilution favour ionisation?

Start with the equilibrium established when a weak monoprotic acid dissolves in water:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

Here:

• \(\ce{HA}\) is the weak acid.
• \(\ce{H3O+}\) is the hydronium ion.
• \(\mathrm{A}^{-}\) is the conjugate base.
• The forward reaction represents ionisation.
• The reverse reaction represents the recombination of ions to form acid molecules.

A weak acid does not ionise completely. At equilibrium, most of its particles remain as undissociated \(\ce{HA}\) molecules.

Now imagine adding water to this equilibrium mixture.

The concentrations of all dissolved species immediately decrease. However, the acid is no longer at equilibrium.

The system responds by favouring the forward reaction, producing additional \(\ce{H3O+}\) and \(\mathrm{A}^{-}\) ions.

Why?

Follow the equilibrium quotient

The acid dissociation constant is:

\[ K_a=\frac{[\ce{H3O+}][\mathrm{A}^{-}]}{[\ce{HA}]} \]

At equilibrium, the reaction quotient \(Q\) has the same value as \(K_a\).

Suppose we dilute the solution by a factor of \(D\). Every dissolved concentration initially becomes \(1/D\) of its original value.

Immediately after dilution:

\[ Q_{\text{new}}= \frac{ ([\ce{H3O+}]/D)([\mathrm{A}^{-}]/D) }{ [\ce{HA}]/D } \]

Simplifying:

\[ Q_{\text{new}}=\frac{K_a}{D} \]

Since \(D>1\):

\[ Q_{\text{new}}<K_a \]

The reaction must proceed forwards to restore equilibrium.

More \(\ce{HA}\) ionises, producing additional hydronium and conjugate base ions, until the equilibrium quotient returns to \(K_a\).

This is the precise reason percentage ionisation increases upon dilution.

You can also picture the particle change: the forward reaction converts one dissolved acid molecule into two dissolved ions. Dilution favours the side with more dissolved particles.

Importantly, this shift does not mean that the hydronium concentration returns to its original value. The solution is still much more dilute than before.

02Deriving the percentage ionisation equation

We can now turn the equilibrium explanation into a calculation.

Consider a pure solution of a weak monoprotic acid with an initial concentration \(C\).

Set up an equilibrium table:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

Here, \(x\) is the concentration of acid that ionises, measured in mol L\(^{-1}\).

We assume that:

• The acid is monoprotic.
• No conjugate base or other common ion was initially added.
• The temperature remains constant.
• Water autoionisation is negligible compared with acid ionisation.

Step 1: Write the equilibrium expression

Substituting the equilibrium concentrations into the \(K_a\) expression:

\[ K_a=\frac{x\times x}{C-x} \]

Therefore:

\[ \boxed{K_a=\frac{x^2}{C-x}} \]

This is the exact concentration-based equilibrium expression under our assumptions.

Step 2: Apply the small-\(x\) approximation

For many weak acid solutions, only a small fraction of the acid ionises.

This means:

\[ x\ll C \]

Therefore, we can approximate:

\[ C-x\approx C \]

The equilibrium expression becomes:

\[ K_a\approx\frac{x^2}{C} \]

Rearranging:

\[ \boxed{x\approx\sqrt{K_aC}} \]

This gives us an estimate of the equilibrium hydronium concentration.

However, we haven’t calculated percentage ionisation yet.

Step 3: Convert to percentage ionisation

Percentage ionisation is the fraction of the original acid concentration that has ionised, multiplied by 100.

Therefore:

\[ \%\text{ ionisation}=\frac{x}{C}\times100 \]

Substitute the approximation for \(x\):

\[ \%\text{ ionisation} \approx \frac{\sqrt{K_aC}}{C}\times100 \]

Simplifying:

\[ \boxed{ \%\text{ ionisation} \approx100\sqrt{\frac{K_a}{C}} } \]

This is the key relationship.

Notice that the initial concentration \(C\) is in the denominator.

As \(C\) decreases, the percentage ionisation increases.

What happens when you dilute by a factor of 10?

Suppose the original concentration is \(C\), and we dilute the solution tenfold.

The new concentration is:

\[ C_{\text{new}}=\frac{C}{10} \]

The original percentage ionisation is:

\[ \%\alpha_{\text{original}} =100\sqrt{\frac{K_a}{C}} \]

After dilution:

\[ \%\alpha_{\text{new}} =100\sqrt{\frac{K_a}{C/10}} \]

Therefore:

\[ \%\alpha_{\text{new}} =\sqrt{10}\,\%\alpha_{\text{original}} \]

Since \(\sqrt{10}\approx3.16\), a tenfold dilution increases the percentage ionisation by approximately 3.16 times, provided the small-\(x\) approximation remains valid.

It does not increase it tenfold.

Why does hydronium concentration decrease at the same time?

This is where students often get caught.

We have established that percentage ionisation increases. But the equilibrium hydronium concentration is approximately:

\[ [\ce{H3O+}]\approx\sqrt{K_aC} \]

Unlike percentage ionisation, this expression contains \(C\) in the numerator.

Consequently, when concentration decreases, hydronium concentration also decreases.

For a tenfold dilution:

\[ [\ce{H3O+}]_{\text{new}} = \frac{[\ce{H3O+}]_{\text{original}}}{\sqrt{10}} \]

So two things happen simultaneously:

• A greater percentage of acid molecules ionise.
• The concentration of hydronium ions decreases.

There is no contradiction. One describes a fraction, while the other describes a concentration.

03Ka, pKa, pH, and percentage ionisation

These four quantities are related, but they do not mean the same thing.

Why doesn’t Ka increase?

The acid dissociation constant is determined by the identity of the acid and the temperature.

For ethanoic acid at 25°C:

\[ K_a=1.8\times10^{-5} \]

Diluting ethanoic acid does not change this value.

The concentrations of the individual species change, and the equilibrium position adjusts, but the equilibrium constant remains the same.

This is an important distinction: a weak acid becomes more extensively ionised upon dilution, but it does not become a stronger acid.

What about pKa?

The pKa is defined as:

\[ pK_a=-\log_{10}K_a \]

For ethanoic acid:

\[ pK_a=-\log_{10}(1.8\times10^{-5}) \]

\[ pK_a\approx4.74 \]

The pKa remains approximately 4.74 when the acid is diluted at the same temperature.

It is a logarithmic measure of the acid dissociation constant, not a measurement of the percentage ionised.

A difference of one pKa unit represents a tenfold difference in \(K_a\), not a tenfold difference in percentage ionisation.

Why does pH increase?

For dilute solutions:

\[ pH\approx-\log_{10}[\ce{H3O+}] \]

Since hydronium concentration decreases upon dilution, pH increases.

For example, if hydronium concentration falls from \(1.0\times10^{-3}\) to \(3.16\times10^{-4}\) mol L\(^{-1}\), the pH increases from 3.00 to approximately 3.50.

The solution is less acidic, even though a greater percentage of the weak acid has ionised.

04How could we experimentally test this?

The equilibrium model makes a prediction. We can test it by preparing several concentrations of ethanoic acid and measuring their pH.

A pH measurement allows us to estimate hydronium concentration. Knowing the original acid concentration then allows us to calculate percentage ionisation.

Setting up the investigation

Start with a 0.100 mol L\(^{-1}\) ethanoic acid solution.

Prepare two further solutions using volumetric dilution:

• Transfer 10.00 mL of the original solution into a 100.0 mL volumetric flask and fill to the calibration mark. This produces a 0.0100 mol L\(^{-1}\) solution.
• Transfer 10.00 mL of the 0.0100 mol L\(^{-1}\) solution into another 100.0 mL volumetric flask and fill to the mark. This produces a 0.00100 mol L\(^{-1}\) solution.

Keep the solutions at the same temperature, ideally 25.0°C.

Measuring pH

A universal indicator could show the general pH trend, but its colour changes are too broad for a reliable percentage-ionisation calculation.

A calibrated pH meter is more suitable.

Rinse the electrode with deionised water between samples, blot it gently, and measure each solution at the same temperature.

For dilute solutions, we can use:

\[ [\ce{H3O+}]\approx10^{-pH} \]

We can then estimate:

\[ \%\text{ ionisation} \approx \frac{10^{-pH}}{C}\times100 \]

This assumes that the acid is the main source of hydronium ions and that hydronium activity can be approximated by its concentration.

What should the results look like?

The following is an illustrative set of predicted pH readings, not a record of an actual experiment.

Notice the pattern.

As the acid is diluted:

• The pH increases.
• Hydronium concentration decreases.
• Percentage ionisation increases.

The third solution also illustrates an important limitation. Its estimated ionisation is greater than 5%, so the small-\(x\) approximation should no longer be trusted. An exact equilibrium calculation is more appropriate.

The experiment therefore tests both the predicted trend and the limitations of our mathematical model.

05A useful contrast: what happens when a strong acid is diluted?

The explanation above applies to weak acids that only partially ionise.

A strong acid behaves differently.

Consider hydrochloric acid:

\[ \ce{HCl + H2O -> H3O+ + Cl-} \]

In ordinary dilute aqueous solutions, hydrochloric acid is essentially completely ionised.

Its percentage ionisation is already approximately 100%.

Diluting it tenfold does not meaningfully increase that percentage.

Instead, its hydronium concentration decreases approximately tenfold.

Compare the two cases:

This comparison highlights why acid strength and acid concentration must be treated separately.

08Questions and solutions

Question 1

A monoprotic weak acid, \(\ce{HA}\), has an acid dissociation constant of \(4.0\times10^{-6}\) at 25°C.

A solution is prepared with an initial concentration of 0.0400 mol L\(^{-1}\).

Calculate its percentage ionisation using the small-\(x\) approximation. Justify whether the approximation is valid.

Solution 1

The percentage ionisation is 1.00%, and the small-\(x\) approximation is valid.

Step 1: Write the equilibrium expression.

\[ K_a=\frac{x^2}{C-x} \]

Assuming \(x\ll C\):

\[ x\approx\sqrt{K_aC} \]

Step 2: Calculate the equilibrium hydronium concentration.

\[ x=\sqrt{(4.0\times10^{-6})(0.0400)} \]

\[ x=\sqrt{1.6\times10^{-7}} \]

\[ x=4.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 3: Calculate percentage ionisation.

\[ \%\text{ ionisation} =\frac{4.0\times10^{-4}}{0.0400}\times100 \]

\[ \boxed{1.00\%} \]

Only 1.00% of the original acid concentration has ionised.

Step 4: Check the approximation.

The calculated value of \(x/C\) is 0.0100, or 1.00%.

Since this is below 5%, the approximation is justified.

The tempting mistake would be to assume that every weak acid automatically satisfies the small-\(x\) condition. It is the calculated fraction, not simply the label ‘weak acid’, that determines whether the approximation is suitable.

Question 2

A monoprotic weak acid has an initial concentration of 0.0250 mol L\(^{-1}\) and a measured pH of 3.00 at 25°C.

Assume that water autoionisation is negligible.

The solution is then diluted fourfold.

Calculate:

(a) The acid dissociation constant, \(K_a\).

(b) The percentage ionisation after dilution.

(c) Whether the small-\(x\) approximation is valid for the diluted solution.

Solution 2

The acid dissociation constant is \(4.17\times10^{-5}\), and the diluted solution is approximately 7.84% ionised. The small-\(x\) approximation is no longer valid after dilution.

Step 1: Use the initial pH to calculate hydronium concentration.

\[ [\ce{H3O+}]=10^{-pH} \]

\[ [\ce{H3O+}]=10^{-3.00} \]

\[ x=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Step 2: Calculate \(K_a\) using the exact equilibrium expression.

The equilibrium acid concentration is:

\[ [\ce{HA}]=0.0250-0.00100 \]

\[ [\ce{HA}]=0.0240\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_a=\frac{x^2}{C-x} \]

\[ K_a=\frac{(1.00\times10^{-3})^2}{0.0240} \]

\[ \boxed{K_a=4.17\times10^{-5}} \]

Step 3: Calculate the diluted concentration.

A fourfold dilution gives:

\[ C_2=\frac{0.0250}{4} \]

\[ C_2=0.00625\ \mathrm{mol\,L^{-1}} \]

Step 4: Test the small-\(x\) approximation.

If we initially assume it is valid:

\[ x\approx\sqrt{K_aC_2} \]

\[ x\approx\sqrt{(4.17\times10^{-5})(0.00625)} \]

\[ x\approx5.10\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

This would give:

\[ \%\text{ ionisation}\approx \frac{5.10\times10^{-4}}{0.00625}\times100 \]

\[ \approx8.17\% \]

This exceeds the 5% guideline.

We should therefore solve the exact equilibrium expression.

Step 5: Solve the quadratic equation.

Starting with:

\[ K_a=\frac{x^2}{C_2-x} \]

Rearrange:

\[ x^2+K_ax-K_aC_2=0 \]

Substituting:

\[ x^2+(4.17\times10^{-5})x -(2.60\times10^{-7})=0 \]

Using the positive quadratic root:

\[ x\approx4.90\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 6: Calculate the actual percentage ionisation.

\[ \%\text{ ionisation} = \frac{4.90\times10^{-4}}{0.00625}\times100 \]

\[ \boxed{7.84\%} \]

The diluted acid is more extensively ionised, as expected.

However, the original small-\(x\) approximation would have overestimated the ionisation. This is why checking the approximation matters, particularly as weak acid solutions become more dilute.

Question 3

Two monoprotic weak acids, P and Q, have the following pKa values at 25°C:

• Acid P: \(pK_a=4.20\)
• Acid Q: \(pK_a=5.20\)

Both acids are initially prepared at 0.0500 mol L\(^{-1}\).

(a) Calculate the approximate percentage ionisation of each acid.

(b) Determine how many times greater the percentage ionisation of P is compared with Q.

(c) Calculate the concentration to which acid Q must be diluted so that its percentage ionisation matches that of acid P at its original concentration.

Assume the small-\(x\) approximation is valid.

Solution 3

Acid P is approximately 3.16 times as ionised as acid Q at the same concentration. Acid Q must be diluted to 0.00500 mol L\(^{-1}\) to match the percentage ionisation of P.

Step 1: Convert the pKa values into \(K_a\).

Recall:

\[ K_a=10^{-pK_a} \]

For acid P:

\[ K_{a,P}=10^{-4.20} \]

\[ K_{a,P}=6.31\times10^{-5} \]

For acid Q:

\[ K_{a,Q}=10^{-5.20} \]

\[ K_{a,Q}=6.31\times10^{-6} \]

Notice that acid P has a \(K_a\) ten times greater than acid Q.

Step 2: Calculate the percentage ionisation of acid P.

Using:

\[ \%\text{ ionisation} \approx100\sqrt{\frac{K_a}{C}} \]

\[ =100\sqrt{\frac{6.31\times10^{-5}}{0.0500}} \]

\[ \boxed{3.55\%} \]

Step 3: Calculate the percentage ionisation of acid Q.

\[ \%\text{ ionisation} \approx100\sqrt{\frac{6.31\times10^{-6}}{0.0500}} \]

\[ \boxed{1.12\%} \]

Both values satisfy the small-\(x\) approximation.

Step 4: Compare the percentages.

\[ \frac{\%\alpha_P}{\%\alpha_Q} = \frac{3.55}{1.12} \]

\[ \boxed{3.16} \]

Acid P is approximately 3.16 times as ionised.

This is an important consequence of the square-root relationship. Although its \(K_a\) is ten times larger, its percentage ionisation is only \(\sqrt{10}\) times larger at the same concentration.

Step 5: Find the concentration required for acid Q.

We want acid Q to have the same percentage ionisation as acid P.

Since:

\[ \%\alpha\propto\sqrt{\frac{K_a}{C}} \]

We can equate the two expressions:

\[ \sqrt{\frac{K_{a,P}}{0.0500}} = \sqrt{\frac{K_{a,Q}}{C_Q}} \]

Squaring both sides:

\[ \frac{K_{a,P}}{0.0500} = \frac{K_{a,Q}}{C_Q} \]

Rearranging:

\[ C_Q= 0.0500\frac{K_{a,Q}}{K_{a,P}} \]

\[ C_Q=0.0500\times\frac{1}{10} \]

\[ \boxed{C_Q=0.00500\ \mathrm{mol\,L^{-1}}} \]

Acid Q must be diluted tenfold to achieve the same percentage ionisation as acid P.

This makes sense: acid Q is weaker, but sufficient dilution allows a greater fraction of its molecules to ionise.

Question 4

Two beakers are prepared at 25°C.

Beaker A contains 0.0100 mol L\(^{-1}\) ethanoic acid.

Beaker B contains 0.0100 mol L\(^{-1}\) ethanoic acid and 0.0100 mol L\(^{-1}\) sodium ethanoate.

The acid dissociation constant of ethanoic acid is:

\[ K_a=1.8\times10^{-5} \]

Assume ideal dilute-solution behaviour.

(a) Calculate the approximate hydronium concentration and pH in each beaker.

(b) Calculate the percentage ionisation of the ethanoic acid in beaker A.

(c) Beaker B is diluted tenfold. Predict its new pH, explaining why the result differs from the behaviour of pure ethanoic acid.

Solution 4

Beaker A has a hydronium concentration of approximately \(4.15\times10^{-4}\) mol L\(^{-1}\) and a pH of 3.38. Beaker B has a hydronium concentration of approximately \(1.8\times10^{-5}\) mol L\(^{-1}\) and a pH of 4.74. After tenfold dilution, the pH of beaker B remains approximately 4.74.

The difference comes from the presence of the conjugate base in beaker B.

Step 1: Calculate the hydronium concentration in beaker A.

Beaker A contains pure ethanoic acid, so we can use:

\[ K_a=\frac{x^2}{C-x} \]

Using the positive quadratic root gives:

\[ x\approx4.15\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ [\ce{H3O+}]_A\approx4.15\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 2: Calculate the pH of beaker A.

\[ pH_A=-\log(4.15\times10^{-4}) \]

\[ \boxed{pH_A=3.38} \]

Step 3: Calculate percentage ionisation in beaker A.

\[ \%\text{ ionisation} = \frac{4.15\times10^{-4}}{0.0100}\times100 \]

\[ \boxed{4.15\%} \]

The small-\(x\) approximation would give approximately 4.24%, which is sufficiently close.

Step 4: Calculate hydronium concentration in beaker B.

Sodium ethanoate dissociates to produce sodium ions and ethanoate ions:

\[ \ce{CH3COONa -> Na+ + CH3COO-} \]

Beaker B therefore already contains a significant concentration of the conjugate base, \(\ce{CH3COO-}\).

The equilibrium expression is:

\[ K_a= \frac{[\ce{H3O+}][\ce{CH3COO-}]} {[\ce{CH3COOH}]} \]

Since the acid and conjugate base concentrations are approximately equal:

\[ K_a\approx[\ce{H3O+}] \]

Therefore:

\[ \boxed{ [\ce{H3O+}]_B\approx1.8\times10^{-5} \ \mathrm{mol\,L^{-1}} } \]

Step 5: Calculate the pH of beaker B.

\[ pH_B=-\log(1.8\times10^{-5}) \]

\[ \boxed{pH_B=4.74} \]

Beaker B has a significantly lower hydronium concentration than beaker A.

The additional ethanoate ions suppress the ionisation of ethanoic acid. This is called the common-ion effect.

Step 6: Predict the effect of tenfold dilution on beaker B.

After dilution, both concentrations decrease by a factor of ten:

\[ [\ce{CH3COOH}]\approx0.00100 \]

\[ [\ce{CH3COO-}]\approx0.00100 \]

Their ratio remains approximately one.

Therefore:

\[ K_a\approx[\ce{H3O+}] \]

The new hydronium concentration is still approximately:

\[ 1.8\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Hence:

\[ \boxed{pH\approx4.74} \]

The pH barely changes.

This is different from diluting pure ethanoic acid because beaker B is a buffer containing both a weak acid and its conjugate base. The square-root percentage-ionisation formula derived for a pure weak acid does not apply to this initial mixture.

The important lesson is to check what is already present in the solution before choosing an equilibrium model.

Question 5

A pure solution of a monoprotic weak acid, \(\ce{HA}\), has:

\[ K_a=1.8\times10^{-5} \]

at 25°C.

A fixed quantity of this acid is progressively diluted, first from an analytical concentration of 0.100 mol L\(^{-1}\) to 0.0100 mol L\(^{-1}\), and eventually to 0.0000100 mol L\(^{-1}\).

Two students make different predictions about the final solution.

Student A uses the small-\(x\) approximation:

\[ \%\text{ ionisation} \approx100\sqrt{\frac{K_a}{C}} \]

They calculate a percentage greater than 100% and conclude that the acid must be completely ionised.

Student B argues that the acid must therefore be exactly 100% ionised, meaning that the final hydronium concentration must equal the analytical acid concentration.

Evaluate both predictions.

Ignoring water autoionisation:

(a) Calculate the exact percentage ionisation at the final concentration using the quadratic equation.

(b) Calculate the final pH.

(c) Compare the total moles of acid ionised before and after dilution, assuming the total moles of acid originally present remain unchanged.

(d) Explain how the total number of ionised acid particles can increase while the hydronium concentration decreases. Identify what additional equilibrium must eventually be considered if the solution is diluted much further.

Solution 5

Neither student’s prediction is correct. At the final concentration of \(1.00\times10^{-5}\) mol L\(^{-1}\), the acid is approximately 71.6% ionised, not 100%. Its pH is approximately 5.15.

The key issue is that the small-\(x\) approximation breaks down when the fraction ionised becomes large.

Step 1: Examine the initial solution.

Initially:

\[ C_1=0.100\ \mathrm{mol\,L^{-1}} \]

Using the small-\(x\) approximation:

\[ x_1\approx\sqrt{K_aC_1} \]

\[ x_1=\sqrt{(1.8\times10^{-5})(0.100)} \]

\[ x_1=1.34\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \%\alpha_1 \approx \frac{1.34\times10^{-3}}{0.100}\times100 \]

\[ \boxed{\%\alpha_1\approx1.34\%} \]

The approximation is valid because the ionisation is small.

Step 2: Examine Student A’s prediction.

The final analytical concentration is:

\[ C_2=1.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Applying the small-\(x\) formula:

\[ \%\alpha_2 \approx100\sqrt{ \frac{1.8\times10^{-5}}{1.00\times10^{-5}} } \]

\[ =100\sqrt{1.8} \]

\[ \approx134\% \]

This result is physically impossible.

The problem is not the equilibrium expression. It is the approximation used to derive it.

The approximation assumed:

\[ x\ll C \]

But the calculated \(x\) is actually greater than \(C\).

That would require more acid to ionise than was originally present.

We must return to the exact expression.

Step 3: Solve the exact equilibrium equation.

Start with:

\[ K_a=\frac{x^2}{C-x} \]

Rearrange:

\[ x^2+K_ax-K_aC=0 \]

Substitute the final concentration:

\[ x^2+(1.8\times10^{-5})x -(1.8\times10^{-10})=0 \]

Using the quadratic formula:

\[ x= \frac{-K_a+\sqrt{K_a^2+4K_aC}}{2} \]

Substituting:

\[ x= \frac{ -1.8\times10^{-5} +\sqrt{(1.8\times10^{-5})^2+ 4(1.8\times10^{-5})(1.0\times10^{-5})} }{2} \]

\[ x=7.16\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate the exact percentage ionisation.

\[ \%\text{ ionisation} = \frac{x}{C}\times100 \]

\[ = \frac{7.16\times10^{-6}} {1.00\times10^{-5}}\times100 \]

\[ \boxed{71.6\%} \]

The acid is substantially ionised, but approximately 28.4% remains undissociated.

Student B’s prediction is therefore also incorrect.

A finite \(K_a\) requires some undissociated acid to remain at equilibrium under these conditions.

Step 5: Calculate the final pH.

Using the equilibrium hydronium concentration:

\[ pH=-\log(7.16\times10^{-6}) \]

\[ \boxed{pH=5.15} \]

The solution is still acidic, even though most of the acid molecules have ionised.

Step 6: Compare the total moles of acid ionised.

The percentage ionisation initially was approximately 1.33% when calculated using the exact equilibrium expression.

After dilution, it is approximately 71.6%.

For a fixed total number of moles of acid:

\[ n_{\text{ionised}}=\alpha n_{\text{total}} \]

Therefore, the ratio of ionised moles is:

\[ \frac{n_{\text{ionised, final}}} {n_{\text{ionised, initial}}} = \frac{0.716}{0.01333} \]

\[ \boxed{\approx53.7} \]

Approximately 53.7 times as many acid molecules are ionised after dilution.

This is possible because the fraction ionised has increased substantially.

Step 7: Compare hydronium concentrations.

Initially:

\[ [\ce{H3O+}]_1 \approx1.33\times10^{-3} \]

Finally:

\[ [\ce{H3O+}]_2 \approx7.16\times10^{-6} \]

The ratio is:

\[ \frac{[\ce{H3O+}]_2} {[\ce{H3O+}]_1} \approx0.00537 \]

Therefore, the final hydronium concentration is approximately 186 times smaller.

This is the subtle distinction.

There are more acid molecules ionised in total, but the solution occupies a much larger volume. Consequently, the concentration of hydronium ions per litre is lower.

Step 8: Consider what happens with further dilution.

As the acid becomes increasingly dilute, its percentage ionisation approaches 100% under the simplified acid-only model.

However, we cannot continue using the same approximation indefinitely.

At concentrations approaching the order of \(10^{-7}\) mol L\(^{-1}\), water autoionisation becomes important:

\[ \ce{2H2O <=> H3O+ + OH-} \]

We must then account for both the acid equilibrium and the ionisation of water, including charge balance.

The hydronium concentration can no longer be treated as coming solely from the acid.

This is the final limitation of the simple model: dilution increases the fraction of weak acid ionised, but the acid-only approximation eventually becomes inadequate.

09Where this leads next

Understanding how dilution affects weak acid equilibrium gives you a useful foundation for predicting what happens when another substance is added.

In particular, adding a strong base removes hydronium ions indirectly by reacting with the weak acid, causing further ionisation and producing a conjugate base.

For the next step, see Neutralisation Models for Strong and Weak Acids and Bases and Weak Acid-Strong Base Titrations: HSC Chemistry Guide.

The same equilibrium expression that explains dilution is what allows you to understand the buffer region and calculate pH during a weak acid-strong base titration.