How to Explain Acid Strength Using Equilibrium Rather Than pH Alone

Two clear acid solutions both have a pH of 2.90. One has an initial concentration of \(0.100\ \mathrm{mol\,L^{-1}}\), while the other has an initial concentration of \(0.0100\ \mathrm{mol\,L^{-1}}\).

Which solution contains the stronger acid?

You might be tempted to choose the more concentrated solution. Or perhaps you’d say they must contain acids of equal strength because their pH values are identical.

Neither conclusion is safe.

Acid strength is determined by how readily an acid donates a proton to water, not simply by the pH of its solution. For a weak acid, we describe this behaviour using its acid dissociation constant, \(K_a\).

pH tells you the hydronium ion concentration at equilibrium. To determine acid strength, you also need to know the original acid concentration and how the proton-transfer equilibrium has settled.

01Start with the proton transfer

Before comparing acid strengths, consider what actually happens when an acid dissolves in water.

According to Brønsted-Lowry theory, an acid is a proton donor, while a base is a proton acceptor.

Consider a generic weak acid, HA, reacting with water:

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

The acid molecule, HA, transfers a proton (\(\mathrm{H}^{+}\)) to a water molecule.

Four chemical species are involved:

This produces two conjugate acid-base pairs:

• HA and A⁻
• H₂O and H₃O⁺

A conjugate acid-base pair consists of two species that differ by exactly one proton.

Notice what happens to HA. Once it donates its proton, it becomes A⁻. This is its conjugate base.

The same idea works in reverse. A⁻ can accept a proton from \(\ce{H3O+}\), reforming HA and water.

What does this have to do with acid strength?

Imagine placing a strong acid and a weak acid into separate containers of water.

A strong acid transfers its protons to water essentially completely.

A weak acid transfers only some of its protons. The remaining acid molecules coexist with the hydronium ions and conjugate base ions.

For a weak acid, the reaction reaches a state of dynamic equilibrium.

This means:

• The forward reaction continues producing \(\ce{H3O+}\) and \(\mathrm{A}^{-}\).
• The reverse reaction continues producing HA and water.
• Both reactions occur at equal rates.
• The concentrations of the dissolved species remain constant.

The reaction has not stopped. The two opposing processes are simply balanced.

A stronger weak acid has a greater tendency to donate protons to water. Its equilibrium therefore favours the products to a greater extent under the same conditions.

There is also an important relationship between conjugate pairs:

The stronger an acid, the weaker its conjugate base.

A strong acid forms a conjugate base that is relatively poor at accepting protons. A weak acid forms a conjugate base that has a greater tendency to accept them.

This relationship helps explain why acid strength is fundamentally about proton transfer, rather than the number of acid particles in a container.

02The equilibrium measurement: Ka

We now have a way to describe acid strength more precisely.

The acid dissociation constant, \(K_a\), measures the position of the equilibrium for an acid reacting with water.

For the general reaction:

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

The acid dissociation constant is:

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

The square brackets represent equilibrium concentrations, measured in \(\mathrm{mol\,L^{-1}}\).

Water does not appear in the expression because it is the liquid solvent and its concentration is effectively constant.

A larger \(K_a\) means the equilibrium ratio of products to remaining acid is larger. The acid therefore has a greater tendency to donate protons.

For two weak acids in the same solvent and at the same temperature:

• Larger \(K_a\) means stronger acid.
• Smaller \(K_a\) means weaker acid.

You can also express acid strength using the logarithmic quantity \(pK_a\):

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

This reverses the direction of the comparison.

A small \(pK_a\) indicates a stronger acid.

Remember that \(K_a\) describes equilibrium composition, not how quickly the reaction occurs. Reaction rate and acid strength are different properties.

03Strength, concentration, amount, and pH are different

These four terms often appear together in HSC Chemistry questions. Mixing them up can lead to incorrect explanations, even when your calculations are correct.

Here:

• \(n\) is the amount of acid in moles.
• \(c\) is the concentration in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the solution volume in litres.

Let’s make the difference concrete.

Imagine two containers holding the same weak acid at \(0.100\ \mathrm{mol\,L^{-1}}\).

The first contains \(50.0\ \mathrm{mL}\). The second contains \(200.0\ \mathrm{mL}\).

Their amounts of acid are:

\[ n_1=0.100\times0.0500 =0.00500\ \mathrm{mol} \]

\[ n_2=0.100\times0.200 =0.0200\ \mathrm{mol} \]

The second container contains four times as many moles of acid.

Does that make its acid stronger?

No. Both containers have the same acid concentration, and the acid has the same intrinsic strength. At the same temperature, their equilibrium hydronium concentrations and pH values will also be the same, assuming equivalent conditions.

The larger container simply contains more solution and more total acid.

This distinction is particularly important when interpreting experimental results.

What pH actually tells you

For dilute aqueous solutions, the HSC calculation is:

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

Rearranging:

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

For example, a solution with pH 2.90 has:

\[ [\ce{H3O+}]=10^{-2.90} \]

\[ =1.26\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

This tells you how much hydronium is present per litre at equilibrium.

It does not tell you:

• The initial concentration of acid.
• The total number of moles of acid added.
• The concentration of un-ionised acid remaining.
• The fraction of the original acid that transferred a proton.

Those missing quantities matter when calculating \(K_a\).

Strictly, pH is defined using hydrogen ion activity rather than concentration. For the dilute solutions considered in these HSC calculations, the concentration approximation is suitable.

If you need to revise the measurement itself, see Measuring pH of Acids and Bases: HSC Chemistry Guide.

04How an experiment can distinguish acid strength

Suppose a chemistry student wants to determine which of two weak acids is stronger.

The student places each solution in a separate beaker and measures its pH using a digital pH meter.

The first solution has pH 2.90. The second has pH 3.20.

The student concludes that the first acid is stronger because it has the lower pH.

Is that conclusion justified?

Not yet.

The student has measured the hydronium ion concentrations, but hasn’t controlled or established the original acid concentrations.

The first solution might simply contain more acid per litre.

Improving the experimental design

A better investigation would control the initial acid concentration and temperature.

The apparatus would include:

• Volumetric flasks for preparing solutions of known concentration.
• Clean beakers for holding the samples.
• A calibrated pH meter with a glass electrode.
• Standard buffer solutions for calibration.
• A wash bottle containing deionised water for rinsing the electrode.

The acid solutions are first prepared at known concentrations in separate volumetric flasks.

Measured portions are then transferred into labelled beakers. The glass electrode is immersed in each sample and connected by cable to the pH meter.

The electrode produces an electrical response related to the hydrogen ion activity. The meter converts this response into a pH reading.

After each measurement, the electrode is rinsed with deionised water to prevent contamination between samples.

The temperature should remain constant because equilibrium constants depend on temperature.

What should the experiment show?

If two weak monoprotic acids have the same initial concentration and are measured at the same temperature, the stronger acid should transfer a greater proportion of its protons to water.

This produces a greater equilibrium hydronium concentration and therefore a lower pH.

However, we can go further than simply comparing the readings.

We can calculate the equilibrium concentrations and determine \(K_a\) for each acid.

That gives us a more precise comparison.

Building an equilibrium table

Consider a weak monoprotic acid with initial concentration \(c_0\).

Its reaction with water is:

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

Let \(x\) represent the concentration of acid that ionises.

Assume the initial contribution from water is negligible and no conjugate base is initially present.

The equilibrium concentrations are:

The important relationship is that one HA molecule produces one hydronium ion and one conjugate base ion.

Therefore:

\[ [\ce{H3O+}]=[\mathrm{A}^{-}]=x \]

We can substitute these equilibrium concentrations into the acid dissociation expression:

\[ K_a= \frac{x\times x}{c_0-x} \]

Therefore:

\[ \boxed{K_a=\frac{x^2}{c_0-x}} \]

And because pH gives us the hydronium concentration:

\[ x=10^{-\mathrm{pH}} \]

We now have a method for determining acid strength from pH and initial concentration.

There is one important condition: this table assumes a weak monoprotic acid, negligible water contribution, and no initially added conjugate base. If those conditions change, the equilibrium table must change too.

05Applying the equilibrium method

Consider the original puzzle again.

Two weak monoprotic acids have the following illustrative measurements, taken at \(25.0^\circ\mathrm{C}\).

The pH readings are identical.

But notice that the initial concentrations are very different.

The next step is to calculate the equilibrium concentration of HA in each solution and compare their \(K_a\) values.

06Important limitations and common mistakes

The equilibrium method is powerful, but you need to know when its assumptions apply.

Mistake 1: Assuming a lower pH always means a stronger acid

A lower pH means a greater hydronium ion concentration.

It does not automatically mean the original acid is stronger.

If two weak monoprotic acids have equal initial concentrations and are measured under identical conditions, their pH values can be used to compare their acid strengths.

If their initial concentrations differ, calculate \(K_a\) instead.

Mistake 2: Treating strong acids exactly like weak acids

A strong acid ionises essentially completely in water.

For example:

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

For a dilute hydrochloric acid solution, we can approximate:

\[ [\ce{H3O+}]\approx[\ce{HCl}]_{\text{initial}} \]

The weak-acid equilibrium table is not the appropriate model for this situation because almost no un-ionised HCl remains.

Also, water has a levelling effect: very strong acids transfer their protons to water essentially completely. Their relative strengths cannot be reliably ranked simply by comparing the pH of equally concentrated aqueous solutions.

Mistake 3: Forgetting the conjugate base

The conjugate base is not an irrelevant product.

It is directly involved in the reverse reaction:

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

A stronger acid forms a weaker conjugate base.

For a conjugate acid-base pair, this relationship can also be expressed through their dissociation constants:

\[ K_a\times K_b=K_w \]

At \(25^\circ\mathrm{C}\), \(K_w=1.0\times10^{-14}\).

A larger acid dissociation constant corresponds to a smaller base dissociation constant for its conjugate base.

Mistake 4: Assuming Ka changes when an acid is diluted

Dilution changes the concentrations of the species in solution.

It can also change the fraction of a weak acid that ionises.

However, at constant temperature and in the same solvent, the acid dissociation constant remains unchanged.

The equilibrium position adjusts to restore the same \(K_a\).

This distinction becomes particularly useful when studying buffers and the effect of adding a conjugate base.

Mistake 5: Ignoring the type of acid

The equilibrium table used throughout this explanation assumes a monoprotic acid, meaning one molecule donates one proton.

A polyprotic acid can donate more than one proton, usually in separate stages.

Each stage has its own equilibrium constant, such as \(K_{a1}\) and \(K_{a2}\).

You cannot automatically use the single-proton equilibrium table to describe every stage of a polyprotic acid.

NSW HSC syllabus scope

The Chemistry Stage 6 (2017) syllabus, which applies to HSC cohorts in 2026 and 2027, includes calculating and applying \(K_a\) and \(pK_a\) to distinguish acid strengths.

NESA’s Chemistry 11-12 Syllabus (2025) is scheduled to begin implementation for Year 11 in 2028, with its first HSC examination in 2029.

The equilibrium method in this article is relevant to the existing 2017 course. You can check the transition arrangements through NESA’s Chemistry Stage 6 syllabus information.

07Questions and solutions

The following questions progress from identifying proton transfer to interpreting equilibrium data and evaluating a more complicated acid system.

Question 1

Consider the following equilibrium reaction:

\[ \ce{HCOOH + H2O <=> H3O+ + HCOO-} \]

Identify:

a) The Brønsted-Lowry acid.

b) The Brønsted-Lowry base.

c) The conjugate acid.

d) The conjugate base.

e) Both conjugate acid-base pairs.

Solution 1

The acid is HCOOH, the base is \(\ce{H2O}\), the conjugate acid is \(\ce{H3O+}\), and the conjugate base is \(\ce{HCOO-}\).

HCOOH donates a proton to water, so it acts as the Brønsted-Lowry acid. Water accepts that proton and acts as the base.

After accepting a proton, water becomes \(\ce{H3O+}\), its conjugate acid.

After donating a proton, HCOOH becomes \(\ce{HCOO-}\), its conjugate base.

The two conjugate pairs are:

• \(\ce{HCOOH/HCOO-}\)
• \(\ce{H2O/H3O+}\)

Each pair differs by exactly one proton.

Question 2

A weak monoprotic acid HX is prepared at an initial concentration of \(0.0400\ \mathrm{mol\,L^{-1}}\).

At \(25.0^\circ\mathrm{C}\), its measured pH is 3.15.

Calculate:

a) The equilibrium hydronium ion concentration.

b) The equilibrium concentration of \(\mathrm{X}^{-}\).

c) The equilibrium concentration of HX.

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

Assume the contribution from water is negligible and no conjugate base was initially added.

Solution 2

The acid dissociation constant is approximately \(1.28\times10^{-5}\).

We first convert the measured pH into the hydronium concentration.

Step 1: Calculate the hydronium concentration.

\[ [\ce{H3O+}]=10^{-3.15} \]

\[ =7.08\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 2: Determine the conjugate base concentration.

From the balanced equation:

\[ \ce{HX + H2O <=> H3O+ + X-} \]

One mole of HX produces one mole of \(\mathrm{X}^{-}\).

Therefore:

\[ [\mathrm{X}^{-}]=7.08\times10^{-4} \]

Step 3: Calculate the remaining acid concentration.

The initial acid concentration was \(0.0400\ \mathrm{mol\,L^{-1}}\).

\[ [\ce{HX}]=0.0400-0.000708 \]

\[ =0.0393\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate \(K_a\).

\[ K_a= \frac{[\ce{H3O+}][\mathrm{X}^{-}]}{[\ce{HX}]} \]

\[ K_a= \frac{(7.08\times10^{-4})^2}{0.0393} \]

\[ \boxed{K_a=1.28\times10^{-5}} \]

The acid is weak because it only partially ionises in water.

Its \(K_a\) value can now be compared with that of another acid measured at the same temperature and in the same solvent.

Question 3

A chemist prepares two solutions of the same weak monoprotic acid HX.

Both are measured at \(25.0^\circ\mathrm{C}\).

a) Calculate \(K_a\) for each solution.

b) Determine whether the results are consistent with both solutions containing the same acid.

c) Explain why increasing the initial concentration by a factor of four does not produce a fourfold increase in hydronium concentration.

Solution 3

Both solutions give approximately the same \(K_a\), confirming that their different pH values are consistent with the same acid having the same strength.

Step 1: Calculate \(K_a\) for solution I.

Its hydronium concentration is:

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

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

The remaining acid concentration is:

\[ [\ce{HX}]_I=0.0500-0.00100 \]

\[ =0.0490\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_{a,I}= \frac{(1.00\times10^{-3})^2}{0.0490} \]

\[ \boxed{K_{a,I}=2.04\times10^{-5}} \]

Step 2: Calculate \(K_a\) for solution II.

Its hydronium concentration is:

\[ [\ce{H3O+}]_{II}=10^{-2.70} \]

\[ =1.995\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

The remaining acid concentration is:

\[ [\ce{HX}]_{II}=0.200-0.001995 \]

\[ =0.1980\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_{a,II}= \frac{(1.995\times10^{-3})^2}{0.1980} \]

\[ \boxed{K_{a,II}=2.01\times10^{-5}} \]

Step 3: Compare the results.

The two values are:

\[ K_{a,I}=2.04\times10^{-5} \]

\[ K_{a,II}=2.01\times10^{-5} \]

These values are very close. The small difference is consistent with rounding of the measured pH values.

Both solutions contain the same acid, so their \(K_a\) values should be the same at the same temperature.

Step 4: Explain the concentration difference.

Solution II has four times the initial acid concentration of solution I.

However, its hydronium concentration is only approximately twice as large:

\[ \frac{1.995\times10^{-3}}{1.00\times10^{-3}} \approx2.00 \]

The reason is that increasing the acid concentration changes the equilibrium composition.

The fraction of acid molecules that ionise is not fixed. The equilibrium adjusts so that the same \(K_a\) is maintained.

This is why you cannot assume hydronium concentration increases in direct proportion to the original concentration of a weak acid.

Question 4

Two students investigate the strength of different weak monoprotic acids, P and Q.

Their experimental data are shown below.

Both solutions are at \(25.0^\circ\mathrm{C}\).

One student claims that P is stronger because it has a lower pH.

The other argues that Q must contain a greater amount of acid because its sample volume is larger.

Evaluate both claims.

Calculate:

a) The initial amount of acid in each sample.

b) The acid dissociation constant for P and Q.

c) Which acid is stronger, and approximately how many times larger is its \(K_a\)?

Solution 4

Acid P is stronger, with an acid dissociation constant approximately twice that of acid Q. However, both samples contain the same initial number of moles of acid.

Step 1: Calculate the amount of acid in each sample.

Use:

\[ n=cV \]

Convert the volumes into litres.

For P:

\[ n_P=0.100\times0.0500 \]

\[ \boxed{n_P=0.00500\ \mathrm{mol}} \]

For Q:

\[ n_Q=0.0500\times0.100 \]

\[ \boxed{n_Q=0.00500\ \mathrm{mol}} \]

Both samples contain the same amount of acid.

The larger volume of Q does not mean it contains more acid because its concentration is half that of P.

Step 2: Calculate \(K_a\) for P.

First, convert its pH into hydronium concentration:

\[ [\ce{H3O+}]_P=10^{-2.90} \]

\[ =1.259\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

The remaining acid concentration is:

\[ [\ce{HP}]=0.100-0.001259 \]

\[ =0.09874\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_{a,P}= \frac{(1.259\times10^{-3})^2}{0.09874} \]

\[ \boxed{K_{a,P}=1.60\times10^{-5}} \]

Step 3: Calculate \(K_a\) for Q.

Its hydronium concentration is:

\[ [\ce{H3O+}]_Q=10^{-3.20} \]

\[ =6.310\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

The remaining acid concentration is:

\[ [\ce{HQ}]=0.0500-0.0006310 \]

\[ =0.04937\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_{a,Q}= \frac{(6.310\times10^{-4})^2}{0.04937} \]

\[ \boxed{K_{a,Q}=8.06\times10^{-6}} \]

Step 4: Compare acid strengths.

\[ \frac{K_{a,P}}{K_{a,Q}} = \frac{1.60\times10^{-5}}{8.06\times10^{-6}} \]

\[ \approx1.99 \]

Acid P has approximately twice the acid dissociation constant of acid Q.

Therefore, P is approximately twice as strong according to \(K_a\).

The first student’s conclusion happens to be correct, but the reasoning is incomplete. A lower pH alone would not establish greater acid strength because the initial concentrations are different.

The second student is also incorrect. The larger sample volume does not mean a greater amount of acid when the concentration is lower.

This question demonstrates why you should calculate the amount, concentration, and equilibrium constant separately rather than treating them as interchangeable.

Question 5

A weak monoprotic acid HA has an initial concentration of \(0.100\ \mathrm{mol\,L^{-1}}\) and a measured pH of 3.00 at \(25.0^\circ\mathrm{C}\).

Its acid dissociation constant is therefore approximately:

\[ K_a=1.01\times10^{-5} \]

A chemist takes two separate \(10.0\ \mathrm{mL}\) samples of this original solution.

The samples are treated as follows:

• Sample D is diluted to a final volume of \(100.0\ \mathrm{mL}\) using deionised water.
• Sample S is diluted to a final volume of \(100.0\ \mathrm{mL}\) using a \(0.0100\ \mathrm{mol\,L^{-1}}\) solution of soluble sodium salt NaA.

Assume that NaA dissociates completely, the volumes are additive, and the solutions behave ideally.

Answer the following:

a) Explain why diluting sample D increases the percentage of HA molecules that ionise, even though its hydronium concentration decreases.

b) Calculate the approximate pH of sample D at equilibrium.

c) Explain why sample S has a different equilibrium composition from sample D, despite both samples having the same final concentration of HA.

d) Calculate the approximate pH of sample S at equilibrium.

e) Explain why neither dilution nor the addition of NaA changes the acid’s \(K_a\).

Solution 5

Sample S has the higher pH, approximately 4.95, while sample D has a pH of approximately 3.50.

The key is that dilution and the addition of a conjugate base affect the equilibrium in different ways. Neither changes the intrinsic acid strength or \(K_a\).

Step 1: Establish the original equilibrium concentrations.

The original solution has:

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

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

From the reaction stoichiometry:

\[ [\mathrm{A}^{-}]=1.00\times10^{-3} \]

The remaining acid concentration is:

\[ [\ce{HA}]=0.100-0.00100 \]

\[ =0.0990\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ K_a= \frac{(0.00100)^2}{0.0990} \]

\[ =1.01\times10^{-5} \]

We will use this same equilibrium constant for both new samples.

Step 2: Analyse sample D after dilution.

The original \(10.0\ \mathrm{mL}\) sample is diluted to \(100.0\ \mathrm{mL}\).

Its initial acid concentration after dilution is:

\[ c_0=0.100\times\frac{10.0}{100.0} \]

\[ =0.0100\ \mathrm{mol\,L^{-1}} \]

The acid is now ten times more dilute.

However, dilution also disturbs the equilibrium.

Immediately after dilution, all dissolved species have been diluted by the same factor. The reaction quotient becomes smaller than \(K_a\), so the forward reaction is favoured.

More HA molecules ionise to restore equilibrium.

This means the percentage of acid molecules that ionise increases, even though the absolute hydronium concentration decreases.

Step 3: Calculate the new hydronium concentration in D.

Use the equilibrium expression:

\[ K_a=\frac{x^2}{c_0-x} \]

Substitute the known values:

\[ 1.01\times10^{-5} = \frac{x^2}{0.0100-x} \]

Rearranging:

\[ x^2+1.01\times10^{-5}x -1.01\times10^{-7}=0 \]

Solving the quadratic gives:

\[ x=3.13\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \mathrm{pH}_D=-\log_{10}(3.13\times10^{-4}) \]

\[ \boxed{\mathrm{pH}_D\approx3.50} \]

The pH has increased from 3.00 to approximately 3.50.

Notice that a tenfold dilution has not increased the pH by exactly one unit. The equilibrium has shifted towards ionisation, producing additional hydronium ions compared with what simple dilution alone would predict.

The percentage ionisation has increased from approximately 1.00% to 3.13%.

Step 4: Analyse sample S.

Sample S also has a final formal concentration of HA equal to:

\[ 0.0100\ \mathrm{mol\,L^{-1}} \]

However, it contains an additional substance: NaA.

The salt dissociates completely:

\[ \ce{NaA -> Na+ + A-} \]

The \(90.0\ \mathrm{mL}\) of added NaA solution produces:

\[ [\mathrm{Na}^{+}]=0.0100\times\frac{90.0}{100.0} \]

\[ =0.00900\ \mathrm{mol\,L^{-1}} \]

It also adds \(0.00900\ \mathrm{mol\,L^{-1}}\) of the conjugate base, \(\mathrm{A}^{-}\).

This is the crucial difference between samples D and S.

Sample S already contains a substantial amount of the product of the acid dissociation reaction.

The added \(\mathrm{A}^{-}\) is a common ion.

According to Le Chatelier’s principle, adding a product favours the reverse reaction:

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

The equilibrium shifts towards HA, reducing the hydronium concentration.

Step 5: Calculate the new hydronium concentration in S.

Let:

\[ [\ce{H3O+}]=x \]

Because the added sodium ion concentration is \(0.00900\ \mathrm{mol\,L^{-1}}\), charge balance gives:

\[ [\mathrm{A}^{-}]=0.00900+x \]

The total concentration of acid-derived species is \(0.0100\ \mathrm{mol\,L^{-1}}\), so:

\[ [\ce{HA}]=0.0100-x \]

Substitute these concentrations into the equilibrium expression:

\[ K_a= \frac{x(0.00900+x)}{0.0100-x} \]

Using the original \(K_a\):

\[ 1.01\times10^{-5} = \frac{x(0.00900+x)}{0.0100-x} \]

Rearranging:

\[ x^2+0.0090101x-1.01\times10^{-7}=0 \]

Solving gives:

\[ x\approx1.12\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \mathrm{pH}_S=-\log_{10}(1.12\times10^{-5}) \]

\[ \boxed{\mathrm{pH}_S\approx4.95} \]

Step 6: Explain why the acid strength has not changed.

Compare the final results:

Both treatments change the equilibrium concentrations, but neither changes \(K_a\).

For sample D, dilution favours further ionisation because the reaction produces more dissolved particles on the product side.

For sample S, the added conjugate base favours the reverse reaction, suppressing ionisation.

The acid itself has not changed.

This is the important distinction: equilibrium composition can change without changing acid strength.

The result also exposes a tempting misconception. A student might assume that more ionisation must always mean a lower pH. But percentage ionisation and absolute hydronium concentration are different quantities. A diluted weak acid can have a greater percentage ionisation while producing a lower hydronium concentration per litre.

This is the bridge to buffer chemistry. A weak acid and its conjugate base can coexist in solution, and changing their relative concentrations changes the equilibrium position and pH without changing the acid’s \(K_a\).

When you explain acid strength in an HSC response, follow the proton, identify the conjugate pairs, and use the equilibrium constant rather than treating pH as a direct measurement of strength.