Why Weak-Acid Strong-Base Equivalence Is Usually Above pH 7

A flask contains ethanoic acid. You add exactly enough sodium hydroxide to react with all the acid. Would a pH meter now read 7.00?

It seems reasonable. The acid and base have neutralised each other, so shouldn’t the solution be neutral?

But at 25°C, the equivalence point of a weak-acid strong-base titration is above pH 7. The strong base converts the weak acid into its conjugate base. That conjugate base then reacts with water, producing hydroxide ions.

The important distinction is that equivalence tells you how many moles have reacted. It doesn’t tell you whether the resulting solution is neutral.

01First, find equivalence without looking at pH

Suppose you have 30.00 mL of \(0.0800\ \mathrm{mol\,L^{-1}}\) ethanoic acid, \(\ce{CH3COOH}\), and titrate it with \(0.1200\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The neutralisation reaction is:

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

One mole of ethanoic acid reacts with one mole of hydroxide ions.

This is a 1:1 mole ratio. It doesn’t matter that ethanoic acid only partially ionises in water. As hydroxide removes protons, more acid ionises and reacts, so the neutralisation proceeds essentially to completion.

The initial moles of acid are calculated using \(n=cV\), where \(n\) is the amount in moles, \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is volume in litres.

\[ \begin{aligned} n&=(0.0800)(0.03000)\\ &=2.40\times10^{-3}\ \mathrm{mol} \end{aligned} \]

At equivalence, the same amount of hydroxide must have been added.

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{2.40\times10^{-3}}{0.1200}\\ &=0.02000\ \mathrm{L}\\ &=20.00\ \mathrm{mL} \end{aligned} \]

So equivalence occurs after exactly 20.00 mL of sodium hydroxide has been delivered, assuming the stated concentrations are accurate.

Notice what we haven’t used: pH.

We know the equivalence volume entirely from reaction stoichiometry.

Now imagine replacing ethanoic acid with 30.00 mL of \(0.0800\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Both samples contain \(2.40\times10^{-3}\) moles of monoprotic acid. Both require 20.00 mL of the same sodium hydroxide solution.

But their equivalence-point pH values are different.

That gives us an experiment worth investigating.

02Testing the prediction: does equivalence mean pH 7?

A simple way to test the original prediction might be to put universal indicator in each flask and add sodium hydroxide until the indicator appears green.

If equivalence really meant neutrality, both flasks should become green at the exact volume needed for neutralisation.

There’s a problem with that design.

Universal indicator gives an approximate pH from colour. It doesn’t directly measure how many moles of acid have reacted. Stopping when the solution looks green assumes that neutrality identifies equivalence, which is exactly the idea we’re supposed to test.

We need to measure pH and identify equivalence independently.

A better experimental arrangement

Use two matched titrations, one with ethanoic acid and one with hydrochloric acid.

For each titration:

• Use a volumetric pipette to transfer 30.00 mL of the \(0.0800\ \mathrm{mol\,L^{-1}}\) acid into a clean conical flask.
• Fill a vertically clamped burette with \(0.1200\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.
• Position the burette tip above the flask so that added solution falls directly into the acid.
• Place the flask on a magnetic stirrer and immerse a calibrated pH electrode in the liquid, clear of the stir bar.
• Add sodium hydroxide in measured increments. Stir, allow the reading to stabilise, and record pH against the volume delivered.

Keep both experiments at 25°C, use the same sodium hydroxide concentration, and take smaller volume increments around 20.00 mL.

The important improvement is that the volume corresponding to equivalence comes from the chemical equation, not from an indicator colour.

The pH electrode measures what happens to the solution at that volume.

What should the revised experiment show?

The table gives calculated predictions for these two titrations. They are illustrative values from an idealised chemical model, not measurements from an experiment.

For ethanoic acid, use \(K_a=1.8\times10^{-5}\). Assume additive volumes, ideal dilute-solution behaviour, and \(K_w=1.00\times10^{-14}\) at 25°C.

Three observations matter.

First, the weak acid begins at a higher pH because it only partially ionises.

Second, both titrations reach equivalence at 20.00 mL, but only the hydrochloric acid solution is neutral at that volume.

Third, once a small excess of sodium hydroxide has been added, both solutions become strongly basic. Their pH values become almost identical.

The model predicting pH 7 for every acid-base equivalence point has failed. The amount of base required is correct, but the predicted pH is not.

03Why the weak-acid flask becomes basic

To understand the result, look at what’s actually in the flask after neutralisation.

When sodium hydroxide reacts with ethanoic acid, it removes a proton from the acid molecule.

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

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

It can accept a proton.

And there’s plenty of water available to donate one.

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

This reaction is called hydrolysis.

An ethanoate ion takes a proton from water, producing an ethanoic acid molecule and a hydroxide ion.

That hydroxide raises the pH.

Think of the proton as a ball being passed between particles.

During neutralisation, hydroxide takes a proton from ethanoic acid. Afterwards, some of the ethanoate ions take protons from water.

The analogy helps you follow which particle holds the proton. It isn’t a one-way sequence, though. Hydrolysis is a reversible equilibrium, with proton transfer occurring in both directions.

Does this mean some sodium hydroxide was left over?

No.

This is the key distinction.

At equivalence, the added hydroxide has been consumed by the neutralisation reaction.

The hydroxide present afterwards is produced by the reaction between ethanoate and water.

The solution is basic because of the conjugate base, not because too much strong base was added.

Hydrolysis also regenerates a small amount of ethanoic acid. That doesn’t mean the original neutralisation was incomplete. The acid is being formed again through a new equilibrium.

In a typical equivalence-point solution, only a small fraction of the ethanoate ions undergo hydrolysis, but that is enough to make the pH noticeably greater than 7.

Why doesn’t hydrochloric acid behave the same way?

For hydrochloric acid and sodium hydroxide, the overall reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

At equivalence, the main dissolved ions are \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\).

Chloride is the conjugate base of a strong acid. It has a negligible tendency to accept a proton from water.

Sodium ions also have negligible acid-base behaviour under these conditions.

Neither ion produces appreciable extra hydroxide or hydronium.

The solution is therefore approximately neutral at 25°C.

So the difference is not how many moles reacted. It’s the acid-base behaviour of the ions remaining after the reaction.

04Calculating how far above pH 7 the equivalence point sits

Knowing that the pH is above 7 is useful. Calculating the actual pH tells us how strongly the conjugate base affects the solution.

For a weak monoprotic acid \(\ce{HA}\), neutralisation produces its conjugate base \(\mathrm{A}^{-}\).

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

The conjugate base has a base dissociation constant, \(K_b\).

For a conjugate acid-base pair:

\[ K_aK_b=K_w \]

Therefore:

\[ K_b=\frac{K_w}{K_a} \]

Here:

• \(K_a\) measures the acid’s tendency to donate a proton.
• \(K_b\) measures its conjugate base’s tendency to accept a proton.
• \(K_w\) is the ion-product constant of water, equal to \(1.00\times10^{-14}\) at 25°C.

A smaller \(K_a\) means a weaker acid and, through this relationship, a stronger conjugate base.

But acid strength isn’t the only factor. The concentration of conjugate base at equivalence also matters.

Suppose the initial concentration of conjugate base after neutralisation is \(C\), in \(\mathrm{mol\,L^{-1}}\).

If hydrolysis produces \(x\ \mathrm{mol\,L^{-1}}\) of hydroxide, the equilibrium relationship is approximately:

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

When only a small fraction hydrolyses, \(x\ll C\), so:

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

Rearranging:

\[ [\ce{OH-}]\approx\sqrt{K_bC} \]

The square brackets indicate concentration in \(\mathrm{mol\,L^{-1}}\).

This approximation assumes hydroxide produced by hydrolysis is much greater than the hydroxide already present from water’s self-ionisation. It works well for the usual moderately concentrated HSC titration examples, but not every possible dilute solution.

Once the hydroxide concentration is known:

\[ \mathrm{pOH}=-\log[\ce{OH-}] \]

At 25°C:

\[ \mathrm{pH}=14.00-\mathrm{pOH} \]

Notice the calculation order: find the equivalence volume, work out the final salt concentration, calculate hydroxide from hydrolysis, and only then calculate pH.

05Why the pH rises sharply just after equivalence

Return to the original ethanoic acid experiment.

The equivalence volume was 20.00 mL.

Imagine that the burette delivers just 0.20 mL too much sodium hydroxide.

That sounds like a tiny error. Will the pH still be around 8.71?

No. The extra hydroxide is now present in stoichiometric excess.

Before equivalence, added hydroxide is consumed by the remaining acid. At equivalence, the conjugate base controls pH through hydrolysis. After equivalence, the excess strong base can dominate.

06What to look for when solving an HSC question

When a question gives you an acid volume, a base concentration, and asks about equivalence-point pH, separate the calculation into two jobs.

First, do the reaction stoichiometry. Determine how many moles of base are required and identify the equivalence volume.

Then identify which particles control the pH.

At the equivalence point of a weak monoprotic acid with NaOH, the dominant acid-base species is the conjugate base produced by neutralisation. Calculate its concentration using the total mixed volume, then use hydrolysis to determine the pH.

If the question instead gives a volume before equivalence, some original weak acid remains alongside its conjugate base. That’s the buffer region. At half-equivalence, the amounts of the two are approximately equal, giving \(\mathrm{pH}\approx\mathrm{p}K_a\), where \(\mathrm{p}K_a=-\log K_a\).

If the added volume exceeds equivalence, calculate the excess strong hydroxide first.

There’s one more useful limitation to remember: the hydrolysis shortcut assumes water’s own ionisation is negligible compared with the hydroxide being produced. If the concentrations become extremely small, or the temperature changes, that assumption needs checking.

A complete treatment also obeys electrical neutrality: the total positive charge in a solution must equal the total negative charge. That becomes useful in the harder questions below.

07Questions and solutions

These questions start with ordinary equivalence-point reasoning and progress towards situations where the usual shortcuts become unreliable.

Assume ideal aqueous solutions and additive volumes unless a question states otherwise.

Question 1

25.0 mL of \(0.0600\ \mathrm{mol\,L^{-1}}\) monoprotic weak acid, \(\ce{HA}\), is titrated with \(0.150\ \mathrm{mol\,L^{-1}}\) sodium hydroxide at 25°C.

Calculate the volume of sodium hydroxide required for equivalence.

State whether the pH at equivalence should be above, below, or equal to 7, and explain using a chemical equation.

Solution 1

The equivalence volume is 10.0 mL, and the equivalence-point pH is above 7 at 25°C.

The neutralisation reaction is:

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

The acid and hydroxide react in a 1:1 mole ratio.

Calculate the original acid moles:

\[ \begin{aligned} n(\ce{HA})&=(0.0600)(0.0250)\\ &=1.50\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The same amount of hydroxide is required.

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{1.50\times10^{-3}}{0.150}\\ &=0.0100\ \mathrm{L}\\ &=10.0\ \mathrm{mL} \end{aligned} \]

At equivalence, the conjugate base \(\mathrm{A}^{-}\) reacts with water:

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

Hydroxide produced by hydrolysis makes the solution basic.

The common mistake is to treat equivalence as a pH measurement. The equivalence volume comes from the mole ratio, while the pH comes from the chemical behaviour of the products.

Question 2

Two unknown monoprotic weak acids, X and Y, are titrated separately with \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide at 25°C.

The experimental conditions are:

The pH values are illustrative experimental data.

A student argues that acid X must be weaker because its initial concentration is higher, so its equivalence solution should contain more conjugate base.

Determine which acid is actually weaker.

Calculate the approximate \(K_a\) for both acids using \(K_w=1.00\times10^{-14}\), and explain why comparing the initial acid concentrations is misleading.

Solution 2

Acid Y is the weaker acid. Its smaller \(K_a\) produces a stronger conjugate base, allowing its more dilute equivalence solution to reach a higher pH.

We need to calculate the final conjugate-base concentrations before interpreting the pH measurements.

First, calculate the original acid amounts.

For X:

\[ n_X=(0.120)(0.0250) =3.00\times10^{-3}\ \mathrm{mol} \]

For Y:

\[ n_Y=(0.0600)(0.0500) =3.00\times10^{-3}\ \mathrm{mol} \]

Both contain exactly the same initial amount of acid.

Therefore, both require:

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{3.00\times10^{-3}}{0.100}\\ &=30.0\ \mathrm{mL} \end{aligned} \]

Next, find the conjugate-base concentration at equivalence.

For X, the total volume is \(25.0+30.0=55.0\ \mathrm{mL}\).

\[ C_X=\frac{3.00\times10^{-3}}{0.0550} =0.0545\ \mathrm{mol\,L^{-1}} \]

For Y, the total volume is \(50.0+30.0=80.0\ \mathrm{mL}\).

\[ C_Y=\frac{3.00\times10^{-3}}{0.0800} =0.0375\ \mathrm{mol\,L^{-1}} \]

So X actually has the more concentrated conjugate-base solution.

Yet Y has the higher pH.

Use the measured pH values to estimate \(K_b\) and \(K_a\).

For X:

\[ \begin{aligned} \mathrm{pOH}_X&=14.00-8.60\\ &=5.40 \end{aligned} \]

\[ [\ce{OH-}]_X=10^{-5.40} =3.98\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

For small hydrolysis:

\[ K_b\approx\frac{[\ce{OH-}]^2}{C} \]

Hence:

\[ \begin{aligned} K_{b,X} &\approx\frac{(3.98\times10^{-6})^2}{0.0545}\\ &=2.91\times10^{-10} \end{aligned} \]

\[ \begin{aligned} K_{a,X} &=\frac{1.00\times10^{-14}}{2.91\times10^{-10}}\\ &=3.44\times10^{-5} \end{aligned} \]

For Y:

\[ \begin{aligned} \mathrm{pOH}_Y&=14.00-8.90\\ &=5.10 \end{aligned} \]

\[ [\ce{OH-}]_Y=10^{-5.10} =7.94\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

\[ \begin{aligned} K_{b,Y} &\approx\frac{(7.94\times10^{-6})^2}{0.0375}\\ &=1.68\times10^{-9} \end{aligned} \]

\[ \begin{aligned} K_{a,Y} &=\frac{1.00\times10^{-14}}{1.68\times10^{-9}}\\ &=5.94\times10^{-6} \end{aligned} \]

Because \(K_{a,Y}<K_{a,X}\), acid Y is weaker.

The surprising result is that Y produces a higher equivalence-point pH despite having a lower conjugate-base concentration.

Its greater conjugate-base strength more than compensates for the dilution.

The student’s mistake is comparing initial concentrations instead of considering both the final salt concentration and the acid’s dissociation constant.

Question 3

A student prepares two identical samples, each containing 25.00 mL of \(0.1000\ \mathrm{mol\,L^{-1}}\) ethanoic acid.

Before titration, the student adds 75.00 mL of pure water to sample B. No water is added to sample A.

Both samples are titrated separately with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide at 25°C.

Use \(K_a=1.8\times10^{-5}\) and \(K_w=1.00\times10^{-14}\).

Determine:

a. The equivalence volume for both samples.

b. The approximate pH of both samples when exactly half the required sodium hydroxide has been added.

c. The approximate equivalence-point pH of both samples.

d. Why the dilution changes one of these pH comparisons but not the other.

Solution 3

Both samples require 25.00 mL of NaOH to reach equivalence. Their half-equivalence pH values are approximately equal at 4.74, but their equivalence-point pH values differ: approximately 8.72 for A and 8.52 for B.

The difference comes from concentration, not from a change in the amount of acid.

Part a: Equivalence volume

Each sample initially contains:

\[ \begin{aligned} n(\ce{CH3COOH}) &=(0.1000)(0.02500)\\ &=2.500\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Adding water does not change this amount.

Therefore:

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{2.500\times10^{-3}}{0.1000}\\ &=0.02500\ \mathrm{L}\\ &=25.00\ \mathrm{mL} \end{aligned} \]

Both equivalence volumes are identical.

Part b: Half-equivalence pH

Half-equivalence occurs after 12.50 mL of sodium hydroxide has been added.

At this stage, approximately half the original ethanoic acid has been converted to ethanoate.

Because the amounts of acid and conjugate base are approximately equal:

\[ \mathrm{pH}\approx\mathrm{p}K_a \]

\[ \begin{aligned} \mathrm{pH} &\approx-\log(1.8\times10^{-5})\\ &=4.74 \end{aligned} \]

Both flasks have approximately this pH.

The added water changes both concentrations, but it does not change the ratio of ethanoate to ethanoic acid in the buffer.

Part c: Equivalence-point pH

At equivalence, the total amount of ethanoate is approximately \(2.500\times10^{-3}\ \mathrm{mol}\).

For sample A:

\[ V_A=25.00+25.00 =50.00\ \mathrm{mL} \]

\[ C_A=\frac{2.500\times10^{-3}}{0.05000} =0.0500\ \mathrm{mol\,L^{-1}} \]

For sample B:

\[ V_B=25.00+75.00+25.00 =125.00\ \mathrm{mL} \]

\[ C_B=\frac{2.500\times10^{-3}}{0.12500} =0.0200\ \mathrm{mol\,L^{-1}} \]

Both contain the same conjugate base, so:

\[ \begin{aligned} K_b&=\frac{1.00\times10^{-14}}{1.8\times10^{-5}}\\ &=5.56\times10^{-10} \end{aligned} \]

For A:

\[ \begin{aligned} [\ce{OH-}]_A &\approx\sqrt{K_bC_A}\\ &=5.27\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

\[ \mathrm{pH}_A =14.00-5.28 =8.72 \]

For B:

\[ \begin{aligned} [\ce{OH-}]_B &\approx\sqrt{K_bC_B}\\ &=3.33\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

\[ \mathrm{pH}_B =14.00-5.48 =8.52 \]

Part d: Explain the different effects of dilution

Half-equivalence pH depends mainly on the ratio between weak acid and conjugate base. Adding water dilutes both species together, so that ratio remains approximately unchanged.

Equivalence-point pH depends on the absolute concentration of the conjugate base, because hydrolysis is the controlling equilibrium.

Sample B is more dilute. It therefore produces a lower hydroxide concentration at equivalence and has a pH closer to 7.

The tempting mistake is assuming that because the equivalence volume stays unchanged, the equivalence-point pH must also stay unchanged.

The first depends on moles. The second also depends on concentration.

Question 4

A student titrates 30.00 mL of \(0.0800\ \mathrm{mol\,L^{-1}}\) ethanoic acid with \(0.1200\ \mathrm{mol\,L^{-1}}\) NaOH at 25°C.

The acid has \(K_a=1.8\times10^{-5}\), and \(K_w=1.00\times10^{-14}\).

During the titration, a calibrated pH meter displays exactly 7.00.

The student immediately declares that equivalence has been reached.

Is this conclusion correct?

Determine the volume of sodium hydroxide at which the solution has pH 7.00, to two decimal places. Explain why this volume is different from the equivalence volume.

Assume ideal dilute-solution behaviour, no other dissolved acids or bases, and additive volumes.

Solution 4

The student is incorrect. The solution reaches pH 7.00 at approximately 19.89 mL of added NaOH, slightly before the 20.00 mL equivalence point.

This is possible because pH 7 and stoichiometric equivalence are different conditions.

First, calculate the true equivalence volume.

\[ \begin{aligned} n_{\mathrm{acid}} &=(0.0800)(0.03000)\\ &=2.40\times10^{-3}\ \mathrm{mol} \end{aligned} \]

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{2.40\times10^{-3}}{0.1200}\\ &=20.00\ \mathrm{mL} \end{aligned} \]

Now determine when the pH is exactly 7.00.

At 25°C:

\[ [\ce{H3O+}]=[\ce{OH-}] =1.00\times10^{-7}\ \mathrm{mol\,L^{-1}} \]

Ethanoic acid and ethanoate are related through:

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

Rearranging:

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

Substitute:

\[ \begin{aligned} \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &=\frac{1.8\times10^{-5}}{1.0\times10^{-7}}\\ &=180 \end{aligned} \]

So, at pH 7, there are 180 moles of ethanoate for every mole of ethanoic acid.

But how much sodium hydroxide does that require?

The solution contains sodium ions, ethanoate ions, hydronium ions, and hydroxide ions.

Electrical neutrality requires:

\[ [\mathrm{Na}^{+}]+[\ce{H3O+}] = [\ce{CH3COO-}]+[\ce{OH-}] \]

At pH 7, the hydronium and hydroxide concentrations are equal. They cancel from the charge balance.

Therefore:

\[ [\mathrm{Na}^{+}]=[\ce{CH3COO-}] \]

This means the moles of sodium hydroxide added equal the moles of ethanoate present.

The total analytical amount of acid remains \(2.40\times10^{-3}\ \mathrm{mol}\), distributed between ethanoic acid and ethanoate.

Because their mole ratio is 180:1:

\[ \begin{aligned} n(\ce{CH3COO-}) &=\frac{180}{181}(2.40\times10^{-3})\\ &=2.3867\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The amount of NaOH added is therefore also \(2.3867\times10^{-3}\ \mathrm{mol}\).

\[ \begin{aligned} V(\ce{NaOH}) &=\frac{2.3867\times10^{-3}}{0.1200}\\ &=0.019889\ \mathrm{L}\\ &=19.89\ \mathrm{mL} \end{aligned} \]

The solution is neutral at 19.89 mL, but equivalence occurs at 20.00 mL.

At pH 7, a small amount of ethanoic acid remains, balancing the conjugate-base equilibrium so that hydronium and hydroxide concentrations are equal.

As the last portion of strong base is added, more ethanoate is produced. At equivalence, hydrolysis makes the solution basic, with pH approximately 8.71.

This result exposes a deeper misconception than simply forgetting hydrolysis.

A pH of 7 doesn’t prove that the neutralisation reaction has reached equivalence. In a weak-acid strong-base titration, the solution can become neutral before equivalence.

The two conditions coincide for an ideal strong-acid strong-base titration at 25°C, but not for this weak acid.

Question 5

Extension challenge: The previous questions used 25°C. What if the experiment is performed at a higher temperature, with extremely dilute solutions?

An investigator prepares 25.00 mL of \(2.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) monoprotic weak acid, \(\ce{HA}\), and titrates it with \(2.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) NaOH at 50°C.

At this temperature:

\[ K_a=1.00\times10^{-4} \]

\[ K_w=5.50\times10^{-14} \]

Assume a closed, carbon-dioxide-free system, ideal dilute-solution behaviour, additive volumes, and no other ions.

Three students make predictions:

• Student A says the equivalence-point pH must be above 7 because the acid is weak.
• Student B uses \([\ce{OH-}]\approx\sqrt{K_bC}\) and concludes that the equivalence solution is neutral.
• Student C argues that the equivalence solution might have pH below 7 while still being basic.

Determine the equivalence volume and calculate the approximate equivalence-point pH.

Which student has the correct physical interpretation?

Explain why the usual hydrolysis approximation is unreliable here, and reconcile your result with the rule that weak-acid strong-base equivalence is above pH 7.

Solution 5

Student C is correct. The solution reaches equivalence after 25.00 mL of NaOH and has an approximate pH of 6.78. Despite being below pH 7, it is basic relative to pure water at 50°C.

Two effects produce this result: the neutral pH changes with temperature, and water’s self-ionisation cannot be ignored in such a dilute solution.

First, locate equivalence using stoichiometry.

The original amount of acid is:

\[ \begin{aligned} n(\ce{HA}) &=(2.00\times10^{-4})(0.02500)\\ &=5.00\times10^{-6}\ \mathrm{mol} \end{aligned} \]

Because the neutralisation ratio is 1:1:

\[ \begin{aligned} V_{\mathrm{eq}} &=\frac{5.00\times10^{-6}}{2.00\times10^{-4}}\\ &=0.02500\ \mathrm{L}\\ &=25.00\ \mathrm{mL} \end{aligned} \]

The total volume is 50.00 mL.

The formal concentration of the resulting conjugate-base salt is:

\[ \begin{aligned} C&=\frac{5.00\times10^{-6}}{0.05000}\\ &=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Next, find the neutral pH at 50°C.

Pure neutral water satisfies:

\[ [\ce{H3O+}]=[\ce{OH-}] \]

Because:

\[ K_w=[\ce{H3O+}][\ce{OH-}] \]

the hydronium concentration in neutral water is:

\[ \begin{aligned} [\ce{H3O+}]_{\mathrm{neutral}} &=\sqrt{5.50\times10^{-14}}\\ &=2.35\times10^{-7}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \mathrm{pH}_{\mathrm{neutral}} =-\log(2.35\times10^{-7}) \approx6.63 \]

At 50°C, neutral water has a pH of approximately 6.63, not 7.00.

So pH below 7 does not automatically mean a solution is acidic at this temperature.

Now examine why the usual hydrolysis approximation fails.

The conjugate base has:

\[ \begin{aligned} K_b&=\frac{5.50\times10^{-14}}{1.00\times10^{-4}}\\ &=5.50\times10^{-10} \end{aligned} \]

Using the familiar approximation gives:

\[ \begin{aligned} [\ce{OH-}] &\approx\sqrt{K_bC}\\ &=\sqrt{5.50\times10^{-14}}\\ &=2.35\times10^{-7}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

That is exactly the hydroxide concentration of neutral water at 50°C.

Student B concludes that the solution is neutral.

But the approximation assumed hydroxide produced by hydrolysis was much greater than hydroxide already present from water.

Here, they’re comparable.

We have used an approximation outside its valid conditions.

Use mass balance and charge balance instead.

Let:

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

and:

\[ b=[\ce{OH-}] \]

Both concentrations are in \(\mathrm{mol\,L^{-1}}\).

At equivalence, the total concentration of acid-derived species is:

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

The sodium concentration is also \(C\).

Electrical neutrality requires:

\[ [\mathrm{Na}^{+}]+h=[\mathrm{A}^{-}]+b \]

Using the mass balance:

\[ b-h=[\ce{HA}] \]

The acid dissociation relationship gives:

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

Combining this with the mass balance gives:

\[ [\ce{HA}]=\frac{Ch}{K_a+h} \]

Because \(h\) is much smaller than \(K_a\):

\[ [\ce{HA}]\approx\frac{Ch}{K_a} \]

In this particular question, \(C=K_a=1.00\times10^{-4}\).

So:

\[ [\ce{HA}]\approx h \]

Return to the charge balance:

\[ b-h\approx h \]

Therefore:

\[ b\approx2h \]

Using \(K_w=hb\):

\[ \begin{aligned} K_w&\approx2h^2\\ h&\approx\sqrt{\frac{K_w}{2}}\\ &=1.66\times10^{-7}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The pH is:

\[ \begin{aligned} \mathrm{pH} &=-\log(1.66\times10^{-7})\\ &\approx6.78 \end{aligned} \]

This approximation is self-consistent because \(1.66\times10^{-7}\) is much smaller than \(K_a=1.00\times10^{-4}\).

Finally:

\[ \begin{aligned} [\ce{OH-}] &=\frac{5.50\times10^{-14}}{1.66\times10^{-7}}\\ &=3.31\times10^{-7}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Compare the two ion concentrations:

\[ [\ce{OH-}]>[\ce{H3O+}] \]

There is more hydroxide than hydronium.

The solution is basic.

This is the limitation hidden inside the usual statement that weak-acid strong-base equivalence is above pH 7.

At 25°C, where neutral water has pH 7.00, the equivalence solution is basic and therefore has pH above 7.

At another temperature, the pH of neutral water can change.

The general principle is more precise: the conjugate base makes the solution basic relative to neutral water at the same temperature. That doesn’t require the numerical pH to exceed 7 under every possible condition.

The deeper lesson is that you must separate three ideas: chemical equivalence, the equilibrium controlling the ion concentrations, and the pH corresponding to neutrality at the stated temperature.

08What this explains about titration measurements

A weak-acid strong-base titration reaches equivalence when the added hydroxide has supplied the stoichiometric amount needed to neutralise the original acid.

Its pH is usually above 7 at 25°C because the conjugate base left behind reacts with water to produce hydroxide.

That distinction tells you how to interpret the steep section of a titration curve, but it also determines which indicator is suitable.

An indicator should change colour within the steep pH transition around equivalence. There’s no reason to insist that its transition must be centred on pH 7.

For the practical difference between a colour change and the stoichiometric point, see Endpoint vs Equivalence Point: The Difference in Titrations.

To connect this equivalence-point reasoning with the earlier buffer region, half-equivalence, and the full titration curve, use Weak Acid-Strong Base Titrations: HSC Chemistry Guide.

The next time a titration question says the acid has been completely neutralised, don’t immediately write pH 7. Work out what particles are left in the flask, then ask what those particles do to the water around them.