Why Strong and Concentrated Mean Different Things for Acids

Put two solutions on the bench: 0.100 mol L⁻¹ hydrochloric acid (\(\ce{HCl}\)) and 0.500 mol L⁻¹ ethanoic acid (\(\ce{CH3COOH}\)).

The ethanoic acid is five times as concentrated. But hydrochloric acid is the stronger acid, and its pH is much lower.

How can that be?

The answer is that acid strength and acid concentration describe two completely different properties. Strength describes how readily an acid transfers a proton to water. Concentration describes how much acid is present in a given volume.

There is also a third quantity to keep separate: the amount of acid, measured in moles.

A concentrated acid is not necessarily strong, and a strong acid is not necessarily concentrated. Understanding this distinction starts with looking at what happens when an acid transfers a proton.

01Three different properties of an acid

Before comparing acids, separate the three questions you might be asking.

Each property answers a different question.

1. Acid strength: how readily does the acid donate a proton?

A strong acid transfers its protons to water essentially completely.

A weak acid transfers only a proportion of its protons. The reaction establishes an equilibrium, with most of the original acid remaining un-ionised.

Acid strength is therefore about the extent of proton transfer, not how many acid particles you started with.

2. Concentration: how much acid is in each litre?

Concentration measures the amount of acid dissolved per unit volume of solution.

The relationship is:

\[ c=\frac{n}{V} \]

Where:

• \(c\) = concentration in mol L⁻¹
• \(n\) = amount of acid in mol
• \(V\) = volume of solution in L

A solution containing 0.500 mol of acid in 1.00 L has a concentration of 0.500 mol L⁻¹.

If you dissolve the same amount in 5.00 L, the concentration becomes 0.100 mol L⁻¹.

The acid has been diluted. Its concentration has changed, but its identity and intrinsic acid strength have not.

3. Amount: how many moles of acid are present altogether?

Imagine two beakers containing the same 0.100 mol L⁻¹ hydrochloric acid solution.

One contains 100 mL, while the other contains 500 mL.

Both have the same concentration and acid strength, but the larger sample contains five times as many moles of acid.

This distinction becomes particularly important in titrations, where the total amount of acid determines how much base is needed for neutralisation.

For a monoprotic weak acid, we can also define its formal concentration, \(C\), as the total concentration of acid-derived species:

\[ C=[\ce{HA}]+[\mathrm{A}^{-}] \]

Some acid molecules remain as \(\ce{HA}\), while others have transferred their protons and become \(\mathrm{A}^{-}\). Both still count towards the original amount of acid.

02Why proton transfer determines acid strength

To understand why one acid is strong and another is weak, consider what happens when an acid meets water.

According to the Brønsted-Lowry model, an acid donates a proton (\(\mathrm{H}^{+}\)), while a base accepts one.

Water acts as the base when it accepts a proton from an acid.

Consider a general acid, \(\ce{HA}\):

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

The acid donates a proton to water, producing a hydronium ion and a conjugate base.

This reaction creates two conjugate acid-base pairs:

• \(\ce{HA}\) and \(\mathrm{A}^{-}\)
• \(\ce{H2O}\) and \(\ce{H3O+}\)

A conjugate acid-base pair differs by exactly one proton.

What happens with a strong acid?

Hydrochloric acid is a strong acid in water.

Its proton transfer is essentially complete:

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

Almost every \(\ce{HCl}\) molecule transfers its proton to water.

The resulting solution contains hydronium and chloride ions, rather than a significant quantity of un-ionised hydrochloric acid.

The chloride ion is the conjugate base of a strong acid, so it has very little tendency to accept a proton back from water.

What happens with a weak acid?

Ethanoic acid behaves differently.

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

Notice the reversible arrow.

Only a small proportion of ethanoic acid molecules transfer their protons to water. Most remain as \(\ce{CH3COOH}\).

The reaction reaches equilibrium, with both reactants and products present.

The acetate ion, \(\ce{CH3COO-}\), is the conjugate base of ethanoic acid.

How do we express acid strength mathematically?

For a weak monoprotic acid, we use the acid dissociation constant, \(K_a\):

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

The square brackets represent equilibrium concentrations.

At a fixed temperature:

• A larger \(K_a\) means the equilibrium favours proton transfer to a greater extent.
• A smaller \(K_a\) means the equilibrium favours the un-ionised acid.
• A larger \(K_a\) therefore indicates a stronger acid.

This gives us a much better way to compare acid strengths than simply looking at their pH values.

For strong acids such as hydrochloric acid, we generally treat ionisation as complete in HSC calculations.

Remember that acid strength is an equilibrium property, not a measure of how quickly an acid reacts.

03Why pH alone cannot tell you acid strength

This is where many students make a mistake.

Suppose you are given two acid solutions:

• Solution X has a pH of 1.5.
• Solution Y has a pH of 2.5.

A student might immediately conclude that X contains the stronger acid.

That conclusion is not necessarily correct.

The pH tells you the hydronium ion concentration, not the intrinsic strength of the acid.

The relationship used in HSC calculations is:

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

A lower pH means a greater hydronium ion concentration.

However, that concentration depends on two things:

1. The initial concentration of the acid.
2. The extent to which the acid transfers its protons to water.

You cannot separate these two effects using pH alone.

How would we design a fair experiment?

Imagine an initial investigation in which a student measures the pH of two unlabelled acid solutions.

The student predicts that whichever solution has the lower pH must contain the stronger acid.

The problem is that the concentrations have not been controlled.

One solution might contain a strong acid at a low concentration, while the other contains a weak acid at a much higher concentration.

The original experiment cannot distinguish these explanations.

To improve the investigation, we need to control the formal concentration.

A fair comparison would follow this process:

1. Prepare hydrochloric acid and ethanoic acid solutions with the same formal concentration.
2. Transfer equal volumes into separate, clean beakers.
3. Keep both solutions at the same temperature.
4. Calibrate the pH meter and measure each solution using the same electrode.
5. Rinse the electrode between measurements to prevent contamination.

The glass electrode responds to the hydronium ion activity in the solution. The meter converts its electrical response into a pH reading.

The important control is that both acids have the same formal concentration.

What would we expect to observe?

Consider the following illustrative equilibrium estimates at 25°C.

These are theoretical HSC-level estimates, not claimed experimental measurements. They use \(K_a=1.8\times10^{-5}\) for ethanoic acid and the usual concentration approximations.

There are two useful comparisons.

Comparison A: HCl versus 0.100 M ethanoic acid

Both solutions have the same formal concentration, but hydrochloric acid produces a much greater hydronium concentration.

This supports the conclusion that hydrochloric acid is stronger.

Comparison B: 0.100 M versus 0.500 M ethanoic acid

Both solutions contain the same weak acid.

The more concentrated solution has a lower pH because it produces a greater hydronium concentration per litre.

However, its acid strength has not changed. It is still ethanoic acid, with the same \(K_a\) at the same temperature.

This is exactly why pH and acid strength must not be treated as interchangeable properties.

A conductivity probe could provide supporting evidence. The hydrochloric acid solution would conduct electricity much more effectively because it contains many more mobile ions. However, conductivity also depends on ion mobility and temperature, so it is not a direct measurement of acid strength.

04An important detail: dilution changes percentage ionisation

There is one subtle point worth understanding.

Although dilution does not change the intrinsic strength of an acid, it can change the percentage of acid molecules that ionise.

Consider ethanoic acid again.

Its equilibrium is:

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

When you dilute a weak acid, its equilibrium shifts towards the products to partially oppose the change.

As a result, a greater percentage of the acid molecules become ionised.

However, the overall hydronium concentration still decreases because the acid has been spread through a larger volume.

For a weak monoprotic acid, when the degree of ionisation is small, we can use:

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

Here, \(C\) is the formal acid concentration.

This equation shows that decreasing the concentration decreases the hydronium concentration, even though the percentage ionisation increases.

For example, consider these two ethanoic acid solutions:

The more dilute solution has a higher percentage ionisation but a lower hydronium concentration.

This is not a contradiction.

Percentage ionisation describes what proportion of the acid molecules transfer their protons. Hydronium concentration describes how many hydronium ions are present per litre.

The key distinction is that \(K_a\) remains constant at a fixed temperature, while the percentage ionisation depends on concentration.

07Questions and solutions

Question 1

Two 30.0 mL samples are prepared:

• Sample A contains 0.150 mol L⁻¹ hydrochloric acid.
• Sample B contains 0.150 mol L⁻¹ ethanoic acid.

(a) Calculate the amount of acid, in moles, in each sample.

(b) Identify which acid is stronger.

(c) Explain whether either solution is more concentrated than the other.

Solution 1

Both samples contain 0.00450 mol of acid, have equal concentrations, and have equal volumes. However, hydrochloric acid is stronger because it ionises essentially completely in water.

Part (a): Calculate the amount of acid.

Use:

\[ n=cV \]

Convert the volume:

\[ 30.0\text{ mL}=0.0300\text{ L} \]

For Sample A:

\[ n=0.150\times0.0300 \]

\[ \boxed{n=0.00450\text{ mol}} \]

For Sample B:

\[ n=0.150\times0.0300 \]

\[ \boxed{n=0.00450\text{ mol}} \]

Part (b): Compare acid strength.

Hydrochloric acid transfers its protons to water essentially completely.

Ethanoic acid only partially ionises, establishing an equilibrium between the acid and its ions.

Therefore, hydrochloric acid is stronger.

Part (c): Compare concentration.

Both solutions have a concentration of 0.150 mol L⁻¹.

Neither is more concentrated.

This question demonstrates that two acids can have identical concentrations and amounts but different strengths.

Question 2

An unknown monoprotic acid, \(\ce{HA}\), has a formal concentration of 0.200 mol L⁻¹.

At 25°C, its measured pH is 2.50.

(a) Calculate the equilibrium hydronium ion concentration.

(b) Calculate the acid dissociation constant, \(K_a\).

(c) Determine the percentage ionisation.

(d) Compare its strength with ethanoic acid, which has \(K_a=1.8\times10^{-5}\).

Explain why the unknown acid can still be classified as weak.

Solution 2

The unknown acid is stronger than ethanoic acid, with \(K_a\approx5.08\times10^{-5}\), but it is still weak because only approximately 1.58% of its molecules ionise.

Part (a): Calculate the hydronium concentration.

Use:

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

Rearrange:

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

Substitute:

\[ [\ce{H3O+}]=10^{-2.50} \]

\[ \boxed{[\ce{H3O+}]=3.16\times10^{-3} \text{ mol L}^{-1}} \]

Part (b): Calculate \(K_a\).

For the equilibrium:

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

The equilibrium concentrations are:

\[ [\ce{H3O+}]=3.16\times10^{-3} \]

\[ [\mathrm{A}^{-}]=3.16\times10^{-3} \]

The concentration of un-ionised acid is:

\[ [\ce{HA}]=0.200-0.00316 \]

\[ [\ce{HA}]=0.19684\text{ mol L}^{-1} \]

Substitute into the \(K_a\) expression:

\[ K_a= \frac{(3.16\times10^{-3})^2}{0.19684} \]

\[ \boxed{K_a=5.08\times10^{-5}} \]

Part (c): Calculate percentage ionisation.

\[ \%\text{ ionisation} = \frac{3.16\times10^{-3}}{0.200} \times100 \]

\[ \boxed{1.58\%} \]

Part (d): Compare acid strength.

Compare the two dissociation constants:

\[ 5.08\times10^{-5}>1.8\times10^{-5} \]

The unknown acid has a larger \(K_a\), so it is stronger than ethanoic acid.

However, only 1.58% of its molecules have ionised.

It is therefore still a weak acid.

The important point is that weak acids do not all have identical strengths. They can have different \(K_a\) values while still undergoing only partial ionisation.

Question 3

A student has 100.0 mL of 0.100 mol L⁻¹ ethanoic acid.

The student adds water until the final solution volume is 1.000 L.

At 25°C, the acid dissociation constant is \(1.8\times10^{-5}\).

Calculate:

(a) The original amount of ethanoic acid.

(b) The new formal concentration after dilution.

(c) The approximate pH before and after dilution.

(d) The percentage ionisation before and after dilution.

Explain why the percentage ionisation increases even though the hydronium concentration decreases.

Solution 3

The amount of ethanoic acid remains 0.0100 mol, while its formal concentration decreases from 0.100 mol L⁻¹ to 0.0100 mol L⁻¹. Its pH increases from approximately 2.87 to 3.37, but its percentage ionisation increases.

Part (a): Calculate the original amount.

Use:

\[ n=cV \]

Convert the initial volume:

\[ 100.0\text{ mL}=0.1000\text{ L} \]

Substitute:

\[ n=0.100\times0.1000 \]

\[ \boxed{n=0.0100\text{ mol}} \]

Adding water does not remove any acid, so this amount remains unchanged.

Part (b): Calculate the new concentration.

The final volume is 1.000 L.

\[ c=\frac{n}{V} \]

\[ c=\frac{0.0100}{1.000} \]

\[ \boxed{c=0.0100\text{ mol L}^{-1}} \]

The concentration has decreased by a factor of ten.

Part (c): Calculate the pH before and after dilution.

For a weak acid where \(x\ll C\):

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

Initially:

\[ [\ce{H3O+}] = \sqrt{(1.8\times10^{-5})(0.100)} \]

\[ =1.34\times10^{-3}\text{ mol L}^{-1} \]

Therefore:

\[ \mathrm{pH}=-\log_{10}(1.34\times10^{-3}) \]

\[ \boxed{\mathrm{pH}=2.87} \]

After dilution:

\[ [\ce{H3O+}] = \sqrt{(1.8\times10^{-5})(0.0100)} \]

\[ =4.24\times10^{-4}\text{ mol L}^{-1} \]

Therefore:

\[ \mathrm{pH}=-\log_{10}(4.24\times10^{-4}) \]

\[ \boxed{\mathrm{pH}=3.37} \]

The pH has increased because the hydronium concentration has decreased.

Part (d): Calculate percentage ionisation.

Initially:

\[ \%\text{ ionisation} = \frac{1.34\times10^{-3}}{0.100} \times100 \]

\[ \boxed{1.34\%} \]

After dilution:

\[ \%\text{ ionisation} = \frac{4.24\times10^{-4}}{0.0100} \times100 \]

\[ \boxed{4.24\%} \]

The percentage ionisation has increased.

This happens because dilution shifts the weak acid equilibrium towards the products, allowing a greater proportion of acid molecules to transfer their protons.

However, the total concentration of hydronium ions still decreases because the solution is much more dilute.

The acid has not become intrinsically stronger. Its \(K_a\) remains unchanged at 25°C.

Question 4

A student separately titrates two 25.0 mL samples:

• Sample A: 0.100 mol L⁻¹ hydrochloric acid.
• Sample B: 0.100 mol L⁻¹ ethanoic acid.

Both are titrated with 0.100 mol L⁻¹ sodium hydroxide.

Assume both acids are monoprotic.

(a) Calculate the volume of sodium hydroxide required to reach the equivalence point for each sample.

(b) Explain why both titrations require the same volume of sodium hydroxide despite the acids having different strengths.

(c) Predict whether the pH at the equivalence point will be equal to, greater than, or approximately equal to 7 for each titration. Explain your reasoning.

Solution 4

Both titrations require 25.0 mL of sodium hydroxide at equivalence. However, the hydrochloric acid titration has an equivalence pH of approximately 7, while the ethanoic acid titration has an equivalence pH greater than 7.

Part (a): Calculate the required volume of sodium hydroxide.

First, calculate the amount of acid in each sample.

\[ n=cV \]

\[ n=0.100\times0.0250 \]

\[ n=0.00250\text{ mol} \]

Both acids are monoprotic, so each mole of acid requires one mole of hydroxide ions for neutralisation.

Therefore:

\[ n(\ce{NaOH})=0.00250\text{ mol} \]

Rearrange the concentration equation:

\[ V=\frac{n}{c} \]

\[ V=\frac{0.00250}{0.100} \]

\[ \boxed{V=0.0250\text{ L}=25.0\text{ mL}} \]

Both titrations require 25.0 mL of sodium hydroxide.

Part (b): Explain why the volumes are equal.

The volume required depends on the total number of moles of acid, not the strength of the acid.

Both samples contain the same amount of monoprotic acid.

Hydrochloric acid is strong, while ethanoic acid is weak. However, the weak acid can still be completely neutralised by a strong base.

The neutralisation reaction for ethanoic acid is:

\[ \ce{CH3COOH + OH- -> CH3COO- + H2O} \]

As hydroxide ions react with the acid, the equilibrium shifts towards further dissociation of ethanoic acid.

Eventually, the acid has been neutralised.

Part (c): Compare the equivalence point pH.

For hydrochloric acid and sodium hydroxide, both the acid and base are strong.

At 25°C, the resulting solution has an equivalence pH of approximately 7.

For ethanoic acid and sodium hydroxide, the acetate ion remains in solution.

Acetate is the conjugate base of a weak acid, so it can accept a proton from water:

\[ \ce{CH3COO- + H2O <=> CH3COOH + OH-} \]

This produces hydroxide ions, making the solution basic.

Therefore:

• HCl and NaOH: equivalence pH approximately 7.
• CH₃COOH and NaOH: equivalence pH greater than 7.

Notice the distinction: both titrations require the same volume of base, but their equivalence point pH values differ.

This is why titration volume and pH tell you different things.

Question 5

Two 25.0 mL acid samples are prepared at 25°C.

Both samples have an initial pH of 3.50.

• Sample A contains hydrochloric acid.
• Sample B contains ethanoic acid.

The acid dissociation constant of ethanoic acid is \(1.8\times10^{-5}\).

Assume ideal solution behaviour, negligible water autoionisation, and one transferable proton per acid molecule.

(a) Calculate the formal concentration of hydrochloric acid.

(b) Calculate the formal concentration of ethanoic acid.

(c) Calculate the amount of acid in each sample.

(d) Calculate the volume of 0.0100 mol L⁻¹ sodium hydroxide required to neutralise each sample completely.

(e) Explain how two samples can have identical pH values but different acid strengths and require very different amounts of sodium hydroxide.

Solution 5

Sample A contains dilute, strong hydrochloric acid, while Sample B contains a more concentrated, weak ethanoic acid. Despite having identical initial pH values, Sample B contains approximately 18.6 times as many moles of acid and requires approximately 18.6 times as much sodium hydroxide for complete neutralisation.

This is a particularly important distinction between hydronium concentration and total acid amount.

Part (a): Calculate the hydrochloric acid concentration.

Hydrochloric acid is strong, so we assume essentially complete ionisation.

Therefore:

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

The initial pH is 3.50.

\[ [\ce{H3O+}]=10^{-3.50} \]

\[ [\ce{H3O+}]=3.16\times10^{-4} \text{ mol L}^{-1} \]

Therefore:

\[ \boxed{C_A=3.16\times10^{-4} \text{ mol L}^{-1}} \]

Part (b): Calculate the ethanoic acid concentration.

Ethanoic acid is weak, so we cannot assume complete ionisation.

Start with:

\[ \ce{CH3COOH + H2O <=> H3O+ + CH3COO-} \]

From the pH reading:

\[ [\ce{H3O+}]=3.16\times10^{-4} \text{ mol L}^{-1} \]

Under the stated assumptions, the acetate concentration is also:

\[ [\ce{CH3COO-}]=3.16\times10^{-4} \text{ mol L}^{-1} \]

Use the acid dissociation expression:

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

Substitute:

\[ 1.8\times10^{-5} = \frac{(3.16\times10^{-4})^2} {[\ce{CH3COOH}]} \]

Rearrange:

\[ [\ce{CH3COOH}] = \frac{(3.16\times10^{-4})^2} {1.8\times10^{-5}} \]

\[ [\ce{CH3COOH}] =5.56\times10^{-3} \text{ mol L}^{-1} \]

This is the equilibrium concentration of un-ionised ethanoic acid.

To find the formal concentration, include the acetate ions as well:

\[ C_B=[\ce{CH3COOH}]+[\ce{CH3COO-}] \]

\[ C_B=5.56\times10^{-3} +3.16\times10^{-4} \]

\[ \boxed{C_B=5.87\times10^{-3} \text{ mol L}^{-1}} \]

The ethanoic acid is approximately 18.6 times as concentrated as the hydrochloric acid.

Part (c): Calculate the amount of acid in each sample.

Both samples have a volume of:

\[ V=25.0\text{ mL}=0.0250\text{ L} \]

For Sample A:

\[ n_A=cV \]

\[ n_A=(3.16\times10^{-4})(0.0250) \]

\[ \boxed{n_A=7.91\times10^{-6}\text{ mol}} \]

For Sample B:

\[ n_B=cV \]

\[ n_B=(5.87\times10^{-3})(0.0250) \]

\[ \boxed{n_B=1.47\times10^{-4}\text{ mol}} \]

Sample B contains approximately 18.6 times as many moles of acid.

Part (d): Calculate the sodium hydroxide volumes.

Both acids are monoprotic, so one mole of sodium hydroxide is required per mole of acid.

The sodium hydroxide concentration is:

\[ c(\ce{NaOH})=0.0100\text{ mol L}^{-1} \]

For Sample A:

\[ V=\frac{n}{c} \]

\[ V_A=\frac{7.91\times10^{-6}}{0.0100} \]

\[ \boxed{V_A=0.791\text{ mL}} \]

For Sample B:

\[ V_B=\frac{1.47\times10^{-4}}{0.0100} \]

\[ \boxed{V_B=14.7\text{ mL}} \]

The ethanoic acid sample requires approximately 18.6 times as much sodium hydroxide.

Part (e): Explain why the samples behave differently.

This is the most important part of the question.

Both samples initially have the same pH because they have the same hydronium ion concentration.

However, the two acids have very different ionisation behaviour.

Hydrochloric acid is strong. Almost all of its acid molecules have transferred their protons to water.

Ethanoic acid is weak. Most of its molecules remain un-ionised as \(\ce{CH3COOH}\).

When sodium hydroxide is added to the ethanoic acid, hydroxide ions react with hydronium ions and the equilibrium shifts towards further dissociation.

The remaining ethanoic acid molecules can continue donating protons.

This means the ethanoic acid solution has a much larger reserve of acid molecules available for neutralisation.

The tempting mistake is to assume that equal pH means equal amounts of acid.

It does not.

The pH measures the hydronium concentration at that particular moment. It does not measure all the acid molecules still capable of donating protons.

This is why the weak ethanoic acid sample can have the same initial pH as the strong hydrochloric acid sample while requiring substantially more sodium hydroxide for complete neutralisation.

08Connecting acid strength to titrations

The distinction between acid strength, concentration, and amount becomes especially useful when interpreting titration curves.

The volume of base required to reach equivalence tells you about the total amount of acid present.

The pH at the equivalence point depends on the acid-base chemistry of the resulting solution, including the behaviour of the conjugate base.

For example, a weak acid reacting with a strong base produces a conjugate base that reacts with water, making the equivalence solution basic.

To build on this reasoning, see Weak Acid-Strong Base Titrations: HSC Chemistry Guide.

Keep three questions separate whenever you analyse an acid:

• How strong is it? Consider proton transfer and \(K_a\).
• How concentrated is it? Calculate moles per litre.
• How much acid is present? Calculate the total number of moles.

Once you can distinguish these three quantities, pH comparisons, equilibrium calculations, and titration questions become much easier to interpret.