How to Prepare a Buffer by Partially Neutralising a Weak Acid

Suppose you add sodium hydroxide to ethanoic acid and neutralise 40% of the original acid. What is the ratio of acetate ions to ethanoic acid molecules in the resulting solution?

A common first guess is 0.40:1. But that isn’t quite right. If 40% of the acid has been converted into acetate, the remaining 60% is still ethanoic acid. The actual ratio is:

\[ \frac{n(\ce{CH3COO-})}{n(\ce{CH3COOH})} =\frac{0.40}{0.60} =0.667 \]

That ratio is the key to preparing a buffer with a particular pH.

To prepare a buffer by partially neutralising a weak acid, add a measured amount of strong base, such as sodium hydroxide, to the weak acid. Stop before all the weak acid has reacted. The reaction converts some of the weak acid into its conjugate base, leaving both species in the solution.

The amount of base you add determines their ratio, which determines the approximate pH of the buffer.

01Why partial neutralisation works

Start with a weak monoprotic acid, represented by \(\ce{HA}\).

In water, it only partially ionises:

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

The equilibrium contains mostly \(\ce{HA}\), with smaller amounts of \(\ce{H3O+}\) and \(\mathrm{A}^{-}\).

Now add sodium hydroxide.

Sodium hydroxide dissociates completely in water:

\[ \ce{NaOH -> Na+ + OH-} \]

The hydroxide ions react with the weak acid. The net ionic equation is:

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

This reaction is treated as essentially complete.

Notice what has happened:

• The weak acid, \(\ce{HA}\), has been consumed.
• Its conjugate base, \(\mathrm{A}^{-}\), has been produced.
• The hydroxide ions have been consumed in the process.

If you add less hydroxide than is needed to neutralise all the acid, some \(\ce{HA}\) remains.

You now have both members of a conjugate acid-base pair:

\[ \boxed{\ce{HA}/\mathrm{A}^{-}} \]

This is the combination needed to make a buffer.

The connection to equilibrium

There is another important detail here.

You might think that sodium hydroxide can only neutralise the small amount of \(\ce{H3O+}\) already present in the weak acid solution.

That prediction would be reasonable if the acid did not undergo equilibrium reactions. But it does.

As \(\ce{OH-}\) removes \(\ce{H3O+}\), the concentration of a product in the weak acid equilibrium decreases. According to Le Chatelier’s principle, the equilibrium shifts to the right.

More \(\ce{HA}\) ionises, producing more \(\ce{H3O+}\) and \(\mathrm{A}^{-}\). The hydroxide continues reacting, and more weak acid is converted into its conjugate base.

The overall result is the net reaction:

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

This is why a weak acid can react with a substantial amount of strong base, even though it was only partially ionised before the base was added.

For a broader explanation of why weak acids can be neutralised essentially completely by strong bases, see Neutralisation Models for Strong and Weak Acids and Bases.

02Calculate how much sodium hydroxide to add

The important calculation has two stages:

1. Use stoichiometry to determine how much weak acid remains and how much conjugate base is produced.
2. Use the acid equilibrium to determine the resulting pH.

Keep these stages separate. It makes buffer preparation much easier.

Step 1: Calculate the initial moles of weak acid

Use:

\[ n=cV \]

where:

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

Call the initial amount of weak acid \(n_0\).

\[ n_0=c_{\ce{HA}}V_{\ce{HA}} \]

Step 2: Calculate the amount of hydroxide added

For sodium hydroxide:

\[ n(\ce{OH-})=c_{\ce{NaOH}}V_{\ce{NaOH}} \]

The reaction between \(\ce{HA}\) and \(\ce{OH-}\) has a 1:1 mole ratio.

Therefore, every mole of hydroxide added:

• consumes one mole of \(\ce{HA}\),
• produces one mole of \(\mathrm{A}^{-}\).

Assuming the hydroxide is less than the initial amount of acid, the resulting mole amounts are:

The small initial ionisation of the weak acid is neglected in this stoichiometric calculation. The table tracks the formal amounts of the acid and its conjugate base.

This is the most important bookkeeping step.

Step 3: Use the buffer equation to find the required ratio

The acid dissociation constant is:

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

Rearranging:

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

Taking negative logarithms gives the Henderson-Hasselbalch equation:

\[ \boxed{ \mathrm{pH} =\mathrm{p}K_a+ \log\left(\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}\right) } \]

Here, \(\mathrm{p}K_a=-\log K_a\).

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

Therefore:

\[ \mathrm{pH} =\mathrm{p}K_a+ \log\left(\frac{n(\mathrm{A}^{-})}{n(\ce{HA})}\right) \]

Substituting the amounts from the neutralisation reaction:

\[ \boxed{ \mathrm{pH} =\mathrm{p}K_a+ \log\left( \frac{n(\ce{OH-})} {n_0-n(\ce{OH-})} \right) } \]

This equation connects the amount of sodium hydroxide added directly to the final buffer pH.

Step 4: Rearrange to calculate the required amount of base

Suppose you know the target pH and the acid’s \(\mathrm{p}K_a\).

First, calculate the required conjugate-base-to-acid ratio:

\[ R=10^{\mathrm{pH}-\mathrm{p}K_a} \]

Then use:

\[ R=\frac{n(\ce{OH-})}{n_0-n(\ce{OH-})} \]

Rearranging:

\[ R(n_0-n(\ce{OH-}))=n(\ce{OH-}) \]

Therefore:

\[ \boxed{ n(\ce{OH-})=\frac{n_0R}{1+R} } \]

Once you know the required moles of hydroxide, calculate the volume of sodium hydroxide:

\[ \boxed{ V_{\ce{NaOH}}= \frac{n(\ce{OH-})}{c_{\ce{NaOH}}} } \]

Remember to use litres when working with concentrations in mol L\(^{-1}\).

There is also a useful way to interpret this result.

The fraction of the original acid neutralised is:

\[ f=\frac{n(\ce{OH-})}{n_0} \]

Therefore:

\[ \boxed{f=\frac{R}{1+R}} \]

This tells you exactly what fraction of the weak acid needs to be converted into its conjugate base.

How much acid should be neutralised?

Consider the following illustrative values.

The key relationship is:

• If the target pH is below the \(\mathrm{p}K_a\), retain more weak acid than conjugate base.
• If the target pH equals the \(\mathrm{p}K_a\), use equal amounts of both.
• If the target pH is above the \(\mathrm{p}K_a\), produce more conjugate base than weak acid.

03How to prepare the buffer in the laboratory

Once the calculations are complete, the practical preparation is straightforward.

1. Select a weak acid with a \(\mathrm{p}K_a\) reasonably close to the required buffer pH.
2. Calculate the initial moles of weak acid.
3. Use the target pH to calculate the required mole ratio of conjugate base to weak acid.
4. Calculate the required moles and volume of sodium hydroxide.
5. Measure the weak acid into a suitable flask.
6. Add the calculated sodium hydroxide slowly while stirring.
7. If a specific final volume is required, transfer the mixture to a volumetric flask and add distilled water to the calibration mark.
8. Mix thoroughly and, if required, measure the final pH using a calibrated pH meter.

Do not simply add half the equivalence volume unless the required pH is equal to the acid’s \(\mathrm{p}K_a\).

Also remember that the volume of sodium hydroxide contributes to the final solution volume. If a final buffer concentration is specified, account for dilution when preparing the solution.

06Why the buffer resists changes in pH

You have now prepared a solution containing both \(\ce{HA}\) and \(\mathrm{A}^{-}\).

But why does this combination resist changes in pH?

The answer is that each component reacts with a different type of addition.

When a small amount of strong acid is added

Suppose you add hydrochloric acid.

It dissociates completely, producing hydronium ions:

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

The conjugate base in your buffer reacts with the added hydronium:

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

The added acid is consumed, producing more weak acid.

This reduces the increase in hydronium concentration and therefore limits the pH decrease.

When a small amount of strong base is added

Now suppose you add sodium hydroxide.

The hydroxide reacts with the weak acid:

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

The added hydroxide is consumed, producing more conjugate base.

This limits the increase in pH.

Think of the two buffer components as separate reserves.

• The weak acid is the reserve that can neutralise added hydroxide.
• The conjugate base is the reserve that can neutralise added hydronium.

However, these reserves are finite. A buffer cannot absorb unlimited acid or base.

07When does the buffer approximation fail?

The Henderson-Hasselbalch equation is extremely useful, but it is not a universal equation for every acid-base mixture.

There are several situations where you must stop and reconsider your chemical model.

1. When one buffer component is almost exhausted

Consider a buffer containing:

• 0.00200 mol of \(\ce{HA}\),
• 0.00800 mol of \(\mathrm{A}^{-}\).

Now add 0.00200 mol of sodium hydroxide.

The hydroxide reacts with all the remaining weak acid:

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

The final mixture contains:

• 0 mol of \(\ce{HA}\),
• 0.01000 mol of \(\mathrm{A}^{-}\).

There is no longer a weak acid/conjugate base pair.

You cannot use Henderson-Hasselbalch because its denominator is zero.

The solution now contains the conjugate base of a weak acid, which reacts with water:

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

You need to calculate the pH using the base equilibrium constant:

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

If even more sodium hydroxide is added, excess strong base controls the pH.

2. When the buffer components are very unbalanced

A practical guide is that an effective buffer usually has a pH within approximately one unit of its \(\mathrm{p}K_a\).

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

Outside this range, one component is much smaller than the other.

The solution may still contain both species, but it has limited capacity to deal with additions that consume the smaller component.

For example, a buffer with a ratio of 100:1 contains very little weak acid relative to its conjugate base. It will have a limited ability to neutralise added hydroxide.

The Henderson-Hasselbalch equation may still provide a mathematical estimate, but calling this a useful buffer would be questionable.

3. When the solution is extremely dilute

The usual buffer equation assumes that the concentrations of the buffer components are large enough for the contribution from water ionisation to be negligible.

If you dilute a buffer sufficiently, this assumption eventually breaks down.

Water itself undergoes ionisation:

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

At very low buffer concentrations, these ions can become significant relative to the buffer components.

The simple Henderson-Hasselbalch approximation then becomes less reliable. A more complete equilibrium calculation must account for water ionisation and electrical charge balance.

4. When you ignore the effect of dilution on buffer capacity

An interesting feature of the Henderson-Hasselbalch equation is that dilution reduces both buffer concentrations by the same factor.

Their ratio therefore stays approximately constant:

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

does not change significantly.

This means dilution usually has little effect on the calculated pH.

However, it does reduce the amount of acid and base available per litre to resist further additions.

Two buffers can have the same pH but very different buffer capacities.

08A contrasting case: Why partially neutralising a strong acid does not make a buffer

A common misconception is that partially neutralising any acid with sodium hydroxide produces a buffer.

That is not true.

Consider hydrochloric acid.

It ionises essentially completely in water:

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

If you partially neutralise it with sodium hydroxide, you consume some of the hydronium ions.

However, the remaining solution contains excess strong acid and chloride ions.

Chloride is the conjugate base of a strong acid. It has negligible ability to react with water to produce hydronium ions.

Therefore, you do not have a useful weak acid/conjugate base pair.

Compare the two situations:

The important requirement is not simply that some acid remains after neutralisation.

You need a weak acid and its conjugate base, both present in appreciable amounts.

09Questions and solutions

Question 1

A student prepares a buffer by mixing 20.0 mL of 0.100 mol L\(^{-1}\) ethanoic acid with 5.00 mL of 0.100 mol L\(^{-1}\) sodium hydroxide.

The \(\mathrm{p}K_a\) of ethanoic acid is 4.76.

Calculate:

a. the final moles of ethanoic acid,

b. the final moles of acetate ions, and

c. the approximate pH of the buffer.

Solution 1

The final mixture contains 0.00150 mol of ethanoic acid and 0.000500 mol of acetate ions, giving a pH of approximately 4.28.

Step 1: Calculate the initial moles of ethanoic acid.

\[ n(\ce{CH3COOH}) =0.100\times0.0200 \]

\[ =0.00200\text{ mol} \]

Step 2: Calculate the moles of hydroxide added.

\[ n(\ce{OH-}) =0.100\times0.00500 \]

\[ =0.000500\text{ mol} \]

Step 3: Use the neutralisation reaction.

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

The reaction consumes 0.000500 mol of ethanoic acid.

Therefore:

\[ n(\ce{CH3COOH})_{\text{remaining}} =0.00200-0.000500 \]

\[ =0.00150\text{ mol} \]

The amount of acetate produced is:

\[ n(\ce{CH3COO-})=0.000500\text{ mol} \]

Step 4: Calculate the pH.

\[ \mathrm{pH} =\mathrm{p}K_a+ \log\left( \frac{n(\ce{CH3COO-})} {n(\ce{CH3COOH})} \right) \]

\[ =4.76+\log\left(\frac{0.000500}{0.00150}\right) \]

\[ =4.76+\log(0.333) \]

\[ \boxed{\mathrm{pH}=4.28} \]

There is more ethanoic acid than acetate in the final solution, so the pH is below the \(\mathrm{p}K_a\).

Question 2

A chemist wants to prepare 100.0 mL of an ethanoic acid buffer with a pH of 5.10.

The available solutions are:

• 40.0 mL of 0.250 mol L\(^{-1}\) ethanoic acid,
• 0.200 mol L\(^{-1}\) sodium hydroxide.

The \(\mathrm{p}K_a\) of ethanoic acid is 4.76.

Calculate:

a. the required mole ratio of acetate to ethanoic acid,

b. the volume of sodium hydroxide required,

c. the volume of water required to reach 100.0 mL, and

d. the final concentrations of both buffer components.

Solution 2

The chemist needs approximately 34.3 mL of sodium hydroxide and 25.7 mL of water. The final concentrations are 0.0686 mol L\(^{-1}\) acetate and 0.0314 mol L\(^{-1}\) ethanoic acid.

Step 1: Calculate the initial moles of ethanoic acid.

\[ n_0=0.250\times0.0400 \]

\[ =0.0100\text{ mol} \]

Step 2: Calculate the required mole ratio.

\[ R=10^{5.10-4.76} \]

\[ R=10^{0.34} \]

\[ R=2.19 \]

The buffer needs approximately 2.19 times as much acetate as ethanoic acid.

Step 3: Calculate the moles of hydroxide required.

\[ n(\ce{OH-})=\frac{n_0R}{1+R} \]

\[ =\frac{0.0100\times2.19}{1+2.19} \]

\[ =0.00686\text{ mol} \]

Step 4: Calculate the sodium hydroxide volume.

\[ V=\frac{0.00686}{0.200} \]

\[ =0.0343\text{ L} \]

\[ \boxed{V=34.3\text{ mL}} \]

Step 5: Calculate the remaining ethanoic acid and acetate.

\[ n(\ce{CH3COOH}) =0.0100-0.00686 \]

\[ =0.00314\text{ mol} \]

And:

\[ n(\ce{CH3COO-})=0.00686\text{ mol} \]

Step 6: Calculate the water volume.

Before dilution, the solution contains:

\[ 40.0+34.3=74.3\text{ mL} \]

Therefore:

\[ V_{\ce{H2O}}=100.0-74.3 \]

\[ \boxed{V_{\ce{H2O}}=25.7\text{ mL}} \]

Step 7: Calculate the final concentrations.

The final volume is 0.1000 L.

For ethanoic acid:

\[ [\ce{CH3COOH}] =\frac{0.00314}{0.1000} \]

\[ =0.0314\text{ mol L}^{-1} \]

For acetate:

\[ [\ce{CH3COO-}] =\frac{0.00686}{0.1000} \]

\[ =0.0686\text{ mol L}^{-1} \]

The ratio is:

\[ \frac{0.0686}{0.0314}=2.19 \]

Therefore:

\[ \mathrm{pH}=4.76+\log(2.19) \]

\[ \boxed{\mathrm{pH}=5.10} \]

The result is consistent with the target pH.

Question 3

A buffer is prepared by partially neutralising 50.0 mL of 0.200 mol L\(^{-1}\) ethanoic acid with 40.0 mL of 0.100 mol L\(^{-1}\) sodium hydroxide.

The \(\mathrm{p}K_a\) of ethanoic acid is 4.76.

A student then adds 10.0 mL of 0.100 mol L\(^{-1}\) hydrochloric acid.

Calculate:

a. the initial moles of ethanoic acid and acetate in the buffer,

b. the pH before adding hydrochloric acid,

c. the pH after adding hydrochloric acid, and

d. the maximum additional moles of hydrochloric acid that could be added before the acetate is completely consumed.

Explain why the pH does not change as dramatically as it would in an unbuffered solution.

Solution 3

After adding hydrochloric acid, the buffer has 0.00700 mol of ethanoic acid and 0.00300 mol of acetate. Its pH is approximately 4.39, and it can consume a further 0.00300 mol of hydrochloric acid before its acetate reserve is exhausted.

Step 1: Calculate the initial moles of ethanoic acid.

\[ n(\ce{HA})=0.200\times0.0500 \]

\[ =0.0100\text{ mol} \]

Step 2: Calculate the moles of sodium hydroxide.

\[ n(\ce{OH-})=0.100\times0.0400 \]

\[ =0.00400\text{ mol} \]

The hydroxide consumes 0.00400 mol of ethanoic acid and produces the same amount of acetate.

Therefore:

\[ n(\ce{HA})=0.0100-0.00400 \]

\[ =0.00600\text{ mol} \]

And:

\[ n(\mathrm{A}^{-})=0.00400\text{ mol} \]

Step 3: Calculate the initial pH.

\[ \mathrm{pH} =4.76+\log\left(\frac{0.00400}{0.00600}\right) \]

\[ =4.76+\log(0.667) \]

\[ \boxed{\mathrm{pH}=4.58} \]

Step 4: Calculate the moles of hydrochloric acid added.

\[ n(\ce{HCl})=0.100\times0.0100 \]

\[ =0.00100\text{ mol} \]

Hydrochloric acid reacts with acetate:

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

Therefore, 0.00100 mol of acetate is consumed, producing an additional 0.00100 mol of ethanoic acid.

The new amounts are:

\[ n(\ce{HA})=0.00600+0.00100 \]

\[ =0.00700\text{ mol} \]

And:

\[ n(\mathrm{A}^{-})=0.00400-0.00100 \]

\[ =0.00300\text{ mol} \]

Step 5: Calculate the new pH.

\[ \mathrm{pH} =4.76+\log\left(\frac{0.00300}{0.00700}\right) \]

\[ =4.76+\log(0.429) \]

\[ \boxed{\mathrm{pH}=4.39} \]

The pH has decreased by approximately 0.19 units.

The added hydrochloric acid has not remained entirely as free hydronium ions. Much of its effect has been absorbed by the acetate ions, which convert into ethanoic acid.

Step 6: Determine the remaining acid reserve.

There are 0.00300 mol of acetate remaining.

Therefore, the maximum additional amount of strong acid that can be consumed stoichiometrically before all the acetate is used is:

\[ \boxed{n(\ce{HCl})=0.00300\text{ mol}} \]

At that point, the buffer has lost its conjugate-base component.

Question 4

Two buffers are prepared using ethanoic acid and sodium ethanoate.

Both have an initial pH of 4.76 and a volume of 100.0 mL.

Their compositions are:

Each buffer receives 0.000500 mol of sodium hydroxide.

Assume the volume change is negligible and use \(K_a=1.74\times10^{-5}\) at \(25^\circ\mathrm{C}\).

a. Calculate the approximate final pH of buffer A.

b. Determine whether buffer B is still a buffer after the addition.

c. Calculate the approximate final pH of buffer B.

d. Explain why the two solutions behaved differently despite having identical initial pH values.

Solution 4

Buffer A remains a buffer and reaches approximately pH 4.85. Buffer B loses all its ethanoic acid and reaches approximately pH 8.38. The difference is their buffer capacity, not their initial pH.

Step 1: Consider the initial compositions.

Both buffers have equal amounts of ethanoic acid and acetate.

Therefore:

\[ \mathrm{pH}=\mathrm{p}K_a=4.76 \]

However, buffer A contains ten times as many moles of each component as buffer B.

Step 2: Calculate the new composition of buffer A.

The added hydroxide is:

\[ 0.000500\text{ mol} \]

It reacts with ethanoic acid.

Remaining acid:

\[ n(\ce{HA})=0.00500-0.000500 \]

\[ =0.00450\text{ mol} \]

New acetate amount:

\[ n(\mathrm{A}^{-})=0.00500+0.000500 \]

\[ =0.00550\text{ mol} \]

Both components remain present, so buffer A is still a buffer.

Step 3: Calculate the new pH of buffer A.

\[ \mathrm{pH} =4.76+\log\left(\frac{0.00550}{0.00450}\right) \]

\[ =4.76+\log(1.222) \]

\[ \boxed{\mathrm{pH}=4.85} \]

The pH has increased by only approximately 0.09 units.

Step 4: Consider buffer B.

Its initial ethanoic acid amount is:

\[ 0.000500\text{ mol} \]

The added hydroxide amount is also:

\[ 0.000500\text{ mol} \]

Therefore, all the ethanoic acid is consumed.

The final mixture contains:

\[ n(\ce{HA})=0 \]

\[ n(\mathrm{A}^{-})=0.00100\text{ mol} \]

Buffer B is no longer a buffer.

Step 5: Calculate the pH of buffer B using base hydrolysis.

The acetate ion reacts with water:

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

First calculate \(K_b\):

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

\[ =\frac{1.0\times10^{-14}}{1.74\times10^{-5}} \]

\[ =5.75\times10^{-10} \]

The acetate concentration is:

\[ [\ce{CH3COO-}] =\frac{0.00100}{0.100} \]

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

Using the small-ionisation approximation:

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

\[ =\sqrt{(5.75\times10^{-10})(0.0100)} \]

\[ =2.40\times10^{-6}\text{ mol L}^{-1} \]

Therefore:

\[ \mathrm{pOH} =-\log(2.40\times10^{-6}) \]

\[ =5.62 \]

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

\[ \mathrm{pH}=14.00-5.62 \]

\[ \boxed{\mathrm{pH}=8.38} \]

Step 6: Explain the difference.

Both buffers started at pH 4.76 because they had equal ratios of conjugate base to weak acid.

But pH and buffer capacity are different properties.

Buffer A had a much larger reserve of ethanoic acid. It could absorb the added hydroxide without exhausting that component.

Buffer B had exactly enough ethanoic acid to react with the added hydroxide. Once that acid was consumed, the buffer system ceased to exist.

This demonstrates why two buffers with identical pH values can have very different abilities to resist pH changes.

Question 5

Two identical flasks each contain 100.0 mL of a buffer with:

• 0.00500 mol of weak acid, \(\ce{HA}\),
• 0.00500 mol of conjugate base, \(\mathrm{A}^{-}\).

The acid has a \(\mathrm{p}K_a\) of 4.76.

Into each flask, the following solutions are added:

• 6.00 mL of 1.00 mol L\(^{-1}\) sodium hydroxide,
• 2.00 mL of 1.00 mol L\(^{-1}\) hydrochloric acid.

The order of addition is different:

• Flask X receives sodium hydroxide first, followed by hydrochloric acid.
• Flask Y receives hydrochloric acid first, followed by sodium hydroxide.

Assume both strong acid and strong base react quantitatively, and ignore activity effects.

a. Calculate the final moles of \(\ce{HA}\) and \(\mathrm{A}^{-}\) in each flask.

b. Determine whether each final solution is a buffer.

c. Calculate the approximate final pH of each solution.

d. Evaluate the statement:

“Once a buffer has been destroyed by adding too much strong base, it cannot become a buffer again.”

Solution 5

Both flasks finish with 0.00100 mol of \(\ce{HA}\) and 0.00900 mol of \(\mathrm{A}^{-}\). Both are buffers with an approximate pH of 5.71, even though flask X temporarily loses its buffer behaviour.

This problem is about tracking the finite amounts of each species through successive reactions.

Step 1: Calculate the amounts of strong acid and strong base added.

For sodium hydroxide:

\[ n(\ce{OH-})=1.00\times0.00600 \]

\[ =0.00600\text{ mol} \]

For hydrochloric acid:

\[ n(\ce{H3O+})=1.00\times0.00200 \]

\[ =0.00200\text{ mol} \]

Each flask receives 0.00600 mol of hydroxide and 0.00200 mol of hydronium.

Step 2: Track flask X, which receives sodium hydroxide first.

Initially:

\[ n(\ce{HA})=0.00500\text{ mol} \]

\[ n(\mathrm{A}^{-})=0.00500\text{ mol} \]

Add 0.00600 mol of hydroxide.

All 0.00500 mol of weak acid is consumed:

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

The resulting amounts are:

\[ n(\ce{HA})=0 \]

\[ n(\mathrm{A}^{-})=0.01000\text{ mol} \]

There is also excess hydroxide:

\[ n(\ce{OH-})_{\text{excess}} =0.00600-0.00500 \]

\[ =0.00100\text{ mol} \]

At this intermediate stage, flask X is not a buffer.

There is no remaining weak acid, and excess strong base is present.

The Henderson-Hasselbalch equation cannot be used here.

Step 3: Add hydrochloric acid to flask X.

The 0.00200 mol of hydronium reacts first with the 0.00100 mol of excess hydroxide:

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

This leaves:

\[ 0.00200-0.00100 =0.00100\text{ mol} \]

of hydronium.

The remaining hydronium reacts with acetate:

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

Therefore, 0.00100 mol of acetate is converted back into weak acid.

The final composition of flask X is:

\[ n(\ce{HA})=0.00100\text{ mol} \]

\[ n(\mathrm{A}^{-})=0.00900\text{ mol} \]

Both components are present again.

Flask X has returned to being a buffer.

Step 4: Track flask Y, which receives hydrochloric acid first.

Initially:

\[ n(\ce{HA})=0.00500\text{ mol} \]

\[ n(\mathrm{A}^{-})=0.00500\text{ mol} \]

The 0.00200 mol of hydronium reacts with acetate:

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

The new composition is:

\[ n(\ce{HA})=0.00500+0.00200 \]

\[ =0.00700\text{ mol} \]

And:

\[ n(\mathrm{A}^{-})=0.00500-0.00200 \]

\[ =0.00300\text{ mol} \]

Step 5: Add sodium hydroxide to flask Y.

The 0.00600 mol of hydroxide reacts with the weak acid:

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

The final amount of weak acid is:

\[ n(\ce{HA})=0.00700-0.00600 \]

\[ =0.00100\text{ mol} \]

The final amount of acetate is:

\[ n(\mathrm{A}^{-})=0.00300+0.00600 \]

\[ =0.00900\text{ mol} \]

Flask Y therefore has exactly the same final composition as flask X.

Step 6: Calculate the final pH of both flasks.

Both contain the same mole ratio:

\[ \frac{n(\mathrm{A}^{-})}{n(\ce{HA})} =\frac{0.00900}{0.00100} =9 \]

Therefore:

\[ \mathrm{pH} =4.76+\log(9) \]

\[ =4.76+0.954 \]

\[ \boxed{\mathrm{pH}=5.71} \]

Both solutions are buffers because they contain appreciable amounts of both the weak acid and its conjugate base.

Step 7: Evaluate the statement.

The statement is incorrect.

A buffer is defined by its final chemical composition, not by the sequence of steps used to prepare it.

Flask X temporarily lost its buffer behaviour when all its weak acid was consumed and excess hydroxide remained. However, the subsequent addition of hydrochloric acid first neutralised the excess hydroxide and then converted some acetate back into ethanoic acid.

The conjugate pair was restored.

This is the important distinction: you cannot use the buffer equation while excess strong acid or base controls the solution, but you can regain a buffer if subsequent reactions produce both conjugate species in sufficient amounts.

The final pH depends on the final mole ratio, provided the buffer approximation is valid.

10What to learn next

Preparing a buffer is only the first part of understanding its behaviour.

You now know how to calculate the initial acid-to-conjugate-base ratio, choose the amount of strong base required, and recognise when one component has been exhausted.

The natural next step is to calculate what happens when an already-prepared buffer receives additional strong acid or strong base.

That requires exactly the same stoichiometric bookkeeping, followed by a new equilibrium calculation.

Continue with How to Calculate Buffer Composition After Adding Strong Acid or Base.