Why a Strong Acid and Its Salt Do Not Form a Buffer

Mix hydrochloric acid with sodium chloride. You now have an acid and a salt containing chloride ions, the conjugate base of HCl. So why doesn’t this mixture act as a buffer?

Predict what happens if you add a small amount of hydrochloric acid. Will the chloride ions absorb the extra acid, keeping the pH relatively stable?

They won’t. HCl is essentially completely ionised in water, and chloride ions are such weak bases that they cannot remove a meaningful amount of added hydronium ions. The salt supplies more chloride, but not a useful supply of a weak base.

This is the crucial distinction: having an acid and its conjugate base present on paper does not automatically give you a functioning buffer. Their relative strengths matter.

01What actually happens when you mix a strong acid and its salt?

Let’s use hydrochloric acid and sodium chloride as our example.

When HCl dissolves in water, it ionises essentially completely:

\[ \ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl^-(aq)} \]

Sodium chloride also dissociates essentially completely:

\[ \ce{NaCl(aq) -> Na+(aq) + Cl^-(aq)} \]

The resulting solution contains:

• \(\ce{H3O+}\), which makes the solution acidic.
• \(\mathrm{Cl}^{-}\), the conjugate base of HCl.
• \(\mathrm{Na}^{+}\), a spectator ion in the acid-base reactions.

At first glance, this looks promising. We have an acid and its conjugate base.

But consider the strength of the two species.

The problem with chloride ions

A conjugate base is formed when an acid donates a proton.

The stronger the acid, the weaker its conjugate base. This relationship is particularly important for strong acids.

HCl is a strong acid. It has a very strong tendency to donate its proton to water. Consequently, \(\mathrm{Cl}^{-}\) has an extremely weak tendency to accept a proton.

We can represent the possible reaction of chloride with water as:

\[ \ce{Cl^-(aq) + H2O(l) <=> HCl(aq) + OH^-(aq)} \]

This equilibrium lies overwhelmingly to the left.

In other words, chloride ions do not react with water to produce an appreciable amount of hydroxide ions.

They also cannot meaningfully remove added hydronium ions through the reverse proton-transfer reaction:

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

The equilibrium strongly favours the left-hand side.

This gives us an important distinction:

Adding sodium chloride increases the chloride concentration, but that doesn’t solve the problem.

The chloride ions are still extremely weak bases, regardless of how many are present.

What happens when acid or base is added?

Consider the two possible disturbances.

Adding more strong acid

The added HCl produces more \(\ce{H3O+}\).

Because chloride cannot meaningfully consume these additional hydronium ions, their concentration increases and the pH decreases.

Adding strong base

The hydroxide ions react with the hydronium ions already present:

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

This removes hydronium ions.

However, there is no useful reserve of weak acid available to replenish them. Once the original hydronium ions have been consumed, excess hydroxide determines the pH.

The mixture therefore cannot effectively resist pH changes in either direction.

02Why a weak acid and its salt form a buffer

Now compare HCl and NaCl with ethanoic acid and sodium ethanoate.

Ethanoic acid, \(\ce{CH3COOH}\), is a weak acid. It only partially ionises in water:

\[ \ce{CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO^-(aq)} \]

Sodium ethanoate dissociates essentially completely:

\[ \ce{CH3COONa(aq) -> Na+(aq) + CH3COO^-(aq)} \]

This mixture contains two important species:

• \(\ce{CH3COOH}\), the weak acid.
• \(\mathrm{CH_{3}COO}^{-}\), its conjugate base.

Unlike the HCl/NaCl mixture, both components can participate meaningfully in acid-base reactions.

What happens when you add strong acid?

Suppose you add hydrochloric acid.

It produces additional hydronium ions, which react with the ethanoate ions:

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

Notice what happens:

• Ethanoate ions are consumed.
• Ethanoic acid is produced.
• Much of the added hydronium is removed.

The equilibrium shifts towards the formation of ethanoic acid.

This is consistent with Le Chatelier’s principle: adding a product to the weak acid equilibrium favours the reverse reaction.

What happens when you add strong base?

Now add sodium hydroxide.

It dissociates to produce hydroxide ions, which react with the weak acid:

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

The ethanoic acid is consumed, and more ethanoate ions are produced.

Removing hydronium through neutralisation causes the weak acid equilibrium to shift towards further ionisation.

The solution therefore resists the increase in pH.

This is why the weak acid and its salt make an effective buffer.

The weak acid provides a reserve to react with added base, while the conjugate base provides a reserve to react with added acid.

Think of these as two finite chemical reserves. Each handles a different type of disturbance.

There is one important limitation: neither reserve is unlimited. Once enough strong acid or base has been added, one component can be consumed completely.

04Buffer capacity: Why even a real buffer eventually fails

Understanding why HCl and NaCl do not form a buffer is only half the story.

Even a correctly prepared weak acid/conjugate base buffer has a limit.

Consider the general weak acid equilibrium:

\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^-(aq)} \]

Here:

• \(\ce{HA}\) is the weak acid.
• \(\mathrm{A}^{-}\) is its conjugate base.
• \(K_a\) is the acid dissociation constant.
• \(pK_a=-\log_{10}(K_a)\).

When strong acid is added, the conjugate base is consumed:

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

When strong base is added, the weak acid is consumed:

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

The amounts of these species are finite.

How do we know when the buffer stops working?

Suppose a buffer initially contains:

• 0.0100 mol of \(\ce{HA}\).
• 0.00600 mol of \(\mathrm{A}^{-}\).

Now add hydrochloric acid.

The relevant comparison is between the moles of hydronium ions added and the initial moles of conjugate base.

The same reasoning works in reverse for strong base. Compare the added hydroxide with the initial amount of weak acid.

This gives you a useful decision rule: calculate the neutralisation reaction before using a buffer equation.

When can you use the Henderson-Hasselbalch equation?

For a functioning weak acid buffer, the Henderson-Hasselbalch equation is:

\[ pH=pK_a+\log_{10} \left(\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}\right) \]

Both buffer components must be present in appreciable amounts.

After adding strong acid or base, calculate their updated mole amounts first.

Because both species are in the same final solution, their concentration ratio is equal to their mole ratio:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} = \frac{n(\mathrm{A}^{-})}{n(\ce{HA})} \]

This makes buffer calculations much easier.

However, the equation is not a universal pH formula.

There are two important situations where you must change your approach:

1. One buffer component has been completely consumed.

If all the conjugate base is consumed, calculate the pH using the weak acid equilibrium.

If all the weak acid is consumed, calculate the pH using the weak base equilibrium of the remaining conjugate base.

2. Strong acid or base remains in excess.

If more strong acid is added than the conjugate base can consume, calculate the concentration of excess hydronium ions.

If more strong base is added than the weak acid can consume, calculate the concentration of excess hydroxide ions.

The excess strong acid or base will generally dominate the pH.

There is also a difference between having both components present and having a useful buffer. A common practical working range is:

\[ pH\approx pK_a\pm1 \]

This corresponds approximately to a conjugate base-to-acid ratio between 0.1 and 10.

It is a useful guide, not an exact boundary at which the Henderson-Hasselbalch equation suddenly stops working.

05How would you test whether the salt actually makes a buffer?

The difference between the two mixtures can be tested using a pH meter.

However, the experiment needs to be designed carefully.

A poor first comparison would be to prepare 0.100 M HCl with 0.100 M NaCl and compare it directly with a 0.100 M acetate buffer.

Their starting pH values would be very different. You would be mixing up initial acidity with resistance to pH change.

A better experiment matches the starting pH and keeps the other conditions constant.

Experimental setup

Prepare two 100.0 mL samples:

Sample A: Acetate buffer

• 0.100 M ethanoic acid.
• 0.100 M sodium ethanoate.

At \(25^\circ\mathrm{C}\), its initial pH is approximately 4.76.

Sample B: Strong acid and its salt

• 0.100 M sodium chloride.
• A small amount of HCl, adjusted to give an initial pH of 4.76.

For the ideal concentration-based model, this corresponds to an initial hydronium concentration of approximately:

\[ [\ce{H3O+}]=10^{-4.76} =1.74\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Place each sample in a separate beaker on a magnetic stirrer.

A calibrated pH electrode is immersed in the solution and connected to a digital pH meter. A burette sits above the beaker so that a measured volume of strong acid or base can be added.

Prediction

If both samples are functioning buffers, adding a small amount of strong acid or base should produce only a small pH change.

But we expect something different from Sample B.

Its chloride ions cannot meaningfully consume added hydronium ions, and it has no weak acid reserve to react with added hydroxide ions.

Expected results

Add 1.00 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) HCl to fresh samples of each solution.

Repeat using fresh samples and the same volume of \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH.

The following values are illustrative predictions from ideal concentration-based calculations, not recorded experimental measurements.

The contrast is substantial.

The acetate buffer changes very little because its two components neutralise much of the added acid or base.

The HCl/NaCl mixture experiences a large pH change because it has neither a useful conjugate base reserve nor a weak acid reserve.

What does the experiment actually measure?

The pH meter measures the pH of the solution. It does not directly measure how many moles of ethanoic acid or ethanoate ions remain.

The chemical equations and mole calculations predict the changing composition. The pH measurements test whether the observed behaviour agrees with those predictions.

Real measurements may differ slightly because of electrode calibration, temperature, and non-ideal solution behaviour.

The important experimental control is that both samples start at the same pH and receive the same amount of acid or base.

This makes the comparison much more useful than simply comparing two solutions with equal starting concentrations.

06An important exception: What about sulfuric acid?

There is one complication worth knowing before applying the rule too broadly.

Sulfuric acid, \(\ce{H2SO4}\), is commonly classified as a strong acid. However, it has two ionisable hydrogen ions, and its two ionisation steps do not behave identically.

The first ionisation is essentially complete:

\[ \ce{H2SO4 + H2O -> H3O+ + HSO4^-} \]

The second ionisation is reversible:

\[ \ce{HSO4^- + H2O <=> H3O+ + SO4^2-} \]

This second equilibrium involves a weak acid and its conjugate base:

• \(\mathrm{HSO_{4}}^{-}\), the hydrogen sulfate ion, is the weak acid.
• \(\mathrm{SO_{4}}^{2-}\), the sulfate ion, is its conjugate base.

Sodium sulfate supplies sulfate ions:

\[ \ce{Na2SO4(aq) -> 2Na+(aq) + SO4^2-(aq)} \]

Consequently, a suitable mixture of sulfuric acid and sodium sulfate can contain an effective \(\mathrm{HSO_{4}}^{-}/\mathrm{SO_{4}}^{2-}\) buffer pair.

This does not contradict the HCl/NaCl explanation.

The important difference is that the buffering pair comes from the second, weaker ionisation step of sulfuric acid. It is not a conventional strong acid/conjugate base pair.

The broader lesson is to examine each ionisation step separately when dealing with a polyprotic acid.

This is an extension to the usual monoprotic strong acid example. Only apply the additional equilibrium calculation when the relevant dissociation constant and assumptions are provided or required.

07Questions and solutions

Use the following data unless otherwise stated:

• \(K_a\) for ethanoic acid = \(1.8\times10^{-5}\).
• \(pK_a\) for ammonium ions = 9.25.
• Temperature = \(25^\circ\mathrm{C}\).
• \(pK_w=14.00\) at \(25^\circ\mathrm{C}\).

Assume ideal concentration-based calculations unless a question states otherwise.

Question 1

A student mixes 25.0 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) HCl with 25.0 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) NaCl.

The student claims that the resulting solution is a buffer because it contains HCl and its conjugate base, \(\mathrm{Cl}^{-}\).

Is the student correct? Explain your reasoning using the relevant chemical equations.

Solution 1

No. The mixture is not a functioning buffer because chloride ions are the extremely weak conjugate base of a strong acid.

Hydrochloric acid ionises essentially completely:

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

Sodium chloride also dissociates completely:

\[ \ce{NaCl -> Na+ + Cl^-} \]

Although chloride is formally the conjugate base of HCl, it has negligible ability to accept additional protons in water.

The mixture therefore lacks an appreciable weak acid/conjugate base equilibrium.

If strong acid is added, the additional hydronium ions remain in solution.

If strong base is added, the hydroxide ions neutralise the existing hydronium ions. There is no useful weak acid reserve to replenish them.

The student’s mistake is assuming that any acid and its conjugate base will form a buffer. The strength of the acid determines whether its conjugate base is sufficiently reactive to provide useful buffering.

Question 2

An ammonia/ammonium buffer contains:

• 0.0150 mol of ammonia, \(\ce{NH3}\).
• 0.0100 mol of ammonium ions, \(\ce{NH4+}\).

A student adds 5.00 mL of \(0.200\ \mathrm{mol\,L^{-1}}\) HCl.

Given that the \(pK_a\) of ammonium ions is 9.25, calculate:

(a) The new mole amounts of ammonia and ammonium ions.

(b) The final pH.

(c) Whether the solution still functions as a buffer.

Solution 2

The final solution contains 0.0140 mol of ammonia and 0.0110 mol of ammonium ions, giving a pH of approximately 9.35. Both buffer components remain, so the solution still functions as a buffer.

Step 1: Calculate the moles of HCl added.

Convert the volume:

\[ V=0.00500\ \mathrm{L} \]

Using \(n=cV\):

\[ \begin{aligned} n(\ce{HCl}) &=0.200(0.00500)\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

Step 2: Write the neutralisation reaction.

\[ \ce{NH3 + H3O+ -> NH4+ + H2O} \]

The added hydronium ions react with ammonia, producing ammonium ions.

Step 3: Calculate the new mole amounts.

Ammonia:

\[ 0.0150-0.00100 =0.0140\ \mathrm{mol} \]

Ammonium ions:

\[ 0.0100+0.00100 =0.0110\ \mathrm{mol} \]

Step 4: Calculate the final pH.

For an ammonia/ammonium buffer:

\[ pH=pK_a+\log_{10} \left(\frac{n(\ce{NH3})}{n(\ce{NH4+})}\right) \]

Substitute the updated amounts:

\[ \begin{aligned} pH&=9.25+\log_{10} \left(\frac{0.0140}{0.0110}\right)\\ &=9.25+\log_{10}(1.2727)\\ &=9.35 \end{aligned} \]

The final pH is approximately 9.35.

The pH decreases because some ammonia has been converted into ammonium ions. The ratio of conjugate base to weak acid has decreased.

Importantly, neither component has been completely consumed. The Henderson-Hasselbalch equation remains appropriate.

Question 3

Two 100.0 mL solutions both have an initial pH of 4.74 at \(25^\circ\mathrm{C}\).

Their compositions are:

Solution A is an acetate buffer. Solution B is an HCl/NaCl mixture.

A student adds 2.00 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) HCl to each solution.

Calculate:

(a) The final pH of Solution A.

(b) The final pH of Solution B.

(c) Explain why the two solutions experience such different pH changes despite having the same initial pH.

Assume ideal concentration-based behaviour and that HCl ionises completely.

Solution 3

Solution A finishes at approximately pH 4.73, while Solution B finishes at approximately pH 2.70. Their different responses arise because only Solution A contains an appreciable weak acid/conjugate base pair.

Part (a): Solution A

Step 1: Calculate the moles of HCl added.

\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=0.100(0.00200)\\ &=0.000200\ \mathrm{mol} \end{aligned} \]

This is 0.200 mmol of hydronium ions.

Step 2: Write the neutralisation reaction.

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

The ethanoate ions are consumed, and ethanoic acid is produced.

Step 3: Calculate the new mole amounts.

Ethanoate ions:

\[ 0.0200-0.000200 =0.0198\ \mathrm{mol} \]

Ethanoic acid:

\[ 0.0200+0.000200 =0.0202\ \mathrm{mol} \]

Step 4: Calculate the final pH.

Using \(pK_a=4.74\):

\[ \begin{aligned} pH&=4.74+\log_{10} \left(\frac{0.0198}{0.0202}\right)\\ &=4.74+\log_{10}(0.9802)\\ &=4.73 \end{aligned} \]

The buffer experiences only a small pH decrease.

Part (b): Solution B

Step 1: Calculate the initial hydronium concentration.

The initial pH is 4.74:

\[ [\ce{H3O+}]_{\text{initial}} =10^{-4.74} \]

\[ =1.82\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Step 2: Calculate the initial moles of hydronium ions.

\[ \begin{aligned} n(\ce{H3O+})_{\text{initial}} &=(1.82\times10^{-5})(0.1000)\\ &=1.82\times10^{-6}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the additional hydronium ions.

\[ \begin{aligned} n(\ce{H3O+})_{\text{added}} &=0.100(0.00200)\\ &=2.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Total hydronium:

\[ \begin{aligned} n_{\text{total}} &=1.82\times10^{-6}+2.00\times10^{-4}\\ &=2.0182\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the final concentration.

The final volume is:

\[ 100.0+2.00=102.0\ \mathrm{mL} \]

\[ V=0.1020\ \mathrm{L} \]

Therefore:

\[ \begin{aligned} [\ce{H3O+}]_{\text{final}} &=\frac{2.0182\times10^{-4}}{0.1020}\\ &=1.98\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 5: Calculate the final pH.

\[ \begin{aligned} pH&=-\log_{10}(1.98\times10^{-3})\\ &=2.70 \end{aligned} \]

Part (c): Explanation

The initial pH values are identical, but the chemical compositions are different.

Solution A contains a substantial amount of ethanoate ions that can react with the added hydronium ions.

Solution B contains chloride ions, which cannot meaningfully absorb the additional hydronium ions.

This question exposes an important misconception: the initial pH of a solution does not tell you its buffer capacity.

You must examine which chemical species are present and how much of each is available to react.

Question 4

A 100.0 mL ethanoic acid/ethanoate buffer contains:

• 0.0120 mol of ethanoic acid.
• 0.00400 mol of ethanoate ions.

A student adds \(0.100\ \mathrm{mol\,L^{-1}}\) HCl.

Determine:

(a) The volume of HCl required to consume all the ethanoate ions.

(b) The concentration of ethanoic acid after this addition.

(c) The final pH, using an appropriate equilibrium calculation.

Given:

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

Assume ideal concentration-based behaviour.

Solution 4

The HCl volume required is 40.0 mL, and the resulting solution has a pH of approximately 2.84. The Henderson-Hasselbalch buffer approximation is no longer appropriate because the original conjugate base has been completely consumed.

Part (a): Calculate the required volume of HCl.

The neutralisation reaction is:

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

The mole ratio is 1:1.

Therefore, 0.00400 mol of HCl is required.

Using \(n=cV\):

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

\[ \begin{aligned} V&=\frac{0.00400}{0.100}\\ &=0.0400\ \mathrm{L}\\ &=40.0\ \mathrm{mL} \end{aligned} \]

Part (b): Calculate the final ethanoic acid concentration.

All 0.00400 mol of ethanoate ions are consumed.

The amount of ethanoic acid becomes:

\[ \begin{aligned} n(\ce{CH3COOH}) &=0.0120+0.00400\\ &=0.0160\ \mathrm{mol} \end{aligned} \]

The final volume is:

\[ 100.0+40.0=140.0\ \mathrm{mL} \]

\[ V=0.1400\ \mathrm{L} \]

Therefore:

\[ \begin{aligned} [\ce{CH3COOH}] &=\frac{0.0160}{0.1400}\\ &=0.114\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Part (c): Calculate the final pH.

At this point, the solution is predominantly a weak acid solution rather than an effective buffer.

Use the weak acid equilibrium:

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

Let \(x\) represent the equilibrium hydronium concentration.

Then:

\[ K_a=\frac{x^2}{0.114-x} \]

Using the small-\(x\) approximation:

\[ 1.8\times10^{-5} \approx\frac{x^2}{0.114} \]

Rearranging:

\[ \begin{aligned} x&\approx\sqrt{(1.8\times10^{-5})(0.114)}\\ &=1.43\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} pH&=-\log_{10}(1.43\times10^{-3})\\ &=2.84 \end{aligned} \]

The final pH is approximately 2.84.

The important detail is that the original ethanoate reserve has been consumed.

A small amount of ethanoate will be regenerated through the weak acid equilibrium, but that does not restore the original buffer capacity.

The tempting mistake would be to substitute zero for the ethanoate amount in the Henderson-Hasselbalch equation. Instead, recognise that the calculation has changed to a weak acid equilibrium problem.

Question 5

Extension: A strong acid with more than one ionisation step

A student claims:

A strong acid and its salt can never form a buffer because the conjugate base of a strong acid is always too weak.

Consider a 1.00 L solution prepared using:

• 0.0100 mol of \(\ce{H2SO4}\).
• 0.100 mol of \(\ce{Na2SO4}\).

At \(25^\circ\mathrm{C}\), assume:

• The first ionisation of sulfuric acid is complete.
• The second ionisation of hydrogen sulfate is reversible.
• \(K_{a2}=1.20\times10^{-2}\).
• Activity effects and water autoionisation are negligible.

The second ionisation is:

\[ \ce{HSO4^- + H2O <=> H3O+ + SO4^2-} \]

Determine:

(a) Whether the resulting solution contains a functioning buffer pair.

(b) The equilibrium concentrations of \(\ce{H3O+}\), \(\mathrm{HSO_{4}}^{-}\), and \(\mathrm{SO_{4}}^{2-}\).

(c) The final pH.

(d) Explain why this result does not contradict the behaviour of HCl and NaCl.

Solution 5

The solution does contain a functioning buffer pair: \(\mathrm{HSO_{4}}^{-}\) and \(\mathrm{SO_{4}}^{2-}\). Its calculated pH is approximately 2.64.

The student’s claim is too broad because sulfuric acid has two ionisation steps, and only its first ionisation is essentially complete.

Step 1: Identify the initial species.

The first ionisation of sulfuric acid produces equal amounts of hydronium and hydrogen sulfate ions:

\[ \ce{H2SO4 + H2O -> H3O+ + HSO4^-} \]

Therefore, initially:

\[ [\ce{H3O+}]=0.0100\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{HSO_{4}}^{-}]=0.0100\ \mathrm{mol\,L^{-1}} \]

Sodium sulfate dissociates:

\[ \ce{Na2SO4 -> 2Na+ + SO4^2-} \]

Therefore:

\[ [\mathrm{SO_{4}}^{2-}]=0.100\ \mathrm{mol\,L^{-1}} \]

Step 2: Determine the direction of the second equilibrium shift.

For the second ionisation:

\[ K_{a2} = \frac{[\ce{H3O+}][\mathrm{SO_{4}}^{2-}]} {[\mathrm{HSO_{4}}^{-}]} \]

Calculate the initial reaction quotient:

\[ Q= \frac{(0.0100)(0.100)}{0.0100} \]

\[ Q=0.100 \]

Since:

\[ Q>K_{a2} \]

the equilibrium shifts to the left.

This means hydronium ions and sulfate ions react to produce hydrogen sulfate ions.

Step 3: Construct the equilibrium expression.

Let \(y\) represent the amount of the reverse reaction in \(\mathrm{mol\,L^{-1}}\).

The equilibrium concentrations become:

Substitute into the equilibrium expression:

\[ 1.20\times10^{-2} = \frac{(0.0100-y)(0.100-y)} {0.0100+y} \]

Step 4: Solve for \(y\).

Rearranging:

\[ (0.0100-y)(0.100-y) = 0.0120(0.0100+y) \]

Expanding and simplifying:

\[ y^2-0.122y+0.000880=0 \]

Solving the quadratic gives two mathematical roots.

Only the smaller root is physically possible because \(y\) cannot exceed the initial hydronium concentration.

\[ y=0.00770\ \mathrm{mol\,L^{-1}} \]

Step 5: Calculate the equilibrium concentrations.

Hydronium ions:

\[ \begin{aligned} [\ce{H3O+}] &=0.0100-0.00770\\ &=0.00230\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Hydrogen sulfate ions:

\[ \begin{aligned} [\mathrm{HSO_{4}}^{-}] &=0.0100+0.00770\\ &=0.0177\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Sulfate ions:

\[ \begin{aligned} [\mathrm{SO_{4}}^{2-}] &=0.100-0.00770\\ &=0.0923\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Both hydrogen sulfate and sulfate ions remain in appreciable amounts.

Step 6: Calculate the final pH.

\[ \begin{aligned} pH&=-\log_{10}(0.00230)\\ &=2.64 \end{aligned} \]

Step 7: Explain why this is a buffer.

The relevant conjugate pair is:

\[ \ce{HSO4^- / SO4^2-} \]

The hydrogen sulfate ions can react with added hydroxide ions:

\[ \ce{HSO4^- + OH^- -> SO4^2- + H2O} \]

The sulfate ions can react with added hydronium ions:

\[ \ce{SO4^2- + H3O+ -> HSO4^- + H2O} \]

Both components are present in sufficient amounts to resist small additions of strong acid or base.

This does not contradict the HCl/NaCl example.

In the HCl/NaCl mixture, chloride is the conjugate base of a strong monoprotic acid and has negligible basicity in water.

In this sulfuric acid example, the first proton is donated essentially completely, but the second ionisation is reversible. That second equilibrium creates the weak acid/conjugate base pair responsible for buffering.

The distinction is subtle but important: you must examine the strength of the particular acid-base pair involved, rather than applying the label strong acid to every ionisation step.

08What to take into your next buffer question

When you see a mixture containing an acid and a salt, don’t immediately assume it is a buffer.

Work through three decisions:

1. Identify the actual acid and base species present after the salt dissociates.
2. Check whether the acid is weak enough and its conjugate base strong enough to participate meaningfully in a reversible equilibrium.
3. Check that both components are present in appreciable amounts after any neutralisation reaction.

For a conventional buffer, you need a weak acid and its conjugate base, or a weak base and its conjugate acid.

For HCl and NaCl, the chloride ions are simply too weak to provide the required buffering action.

Once you can recognise a genuine buffer, the next skill is tracking how its composition changes when strong acid or base is added. Continue with How to Recognise a Buffer from Mixed Solutions and How to Calculate Buffer Composition After Adding Strong Acid or Base.