When Can You Use the Henderson-Hasselbalch Equation?

A buffer can start at pH 4.74, then receive a measured amount of hydrochloric acid. If it contains plenty of ethanoate ions, most of the added acid will be neutralised, and the pH will change only slightly. But what happens if the added acid nearly uses up all the ethanoate?

This is where the Henderson-Hasselbalch equation becomes less reliable.

You can use the Henderson-Hasselbalch equation when a solution contains appreciable amounts of a weak acid and its conjugate base, and the equilibrium changes their concentrations by relatively small amounts. If a strong acid or base has been added, you must first account for the neutralisation reaction. If either buffer component is nearly exhausted, the usual approximation may fail.

There is an important distinction here: the Henderson-Hasselbalch relationship is mathematically exact when equilibrium concentrations are used (assuming ideal behaviour). Its common shortcut calculation uses the amounts remaining after neutralisation as approximations to those equilibrium concentrations. That approximation is what we need to test.

01What the Henderson-Hasselbalch equation actually tells you

Consider a weak acid, represented by \(\mathrm{HA}\), which partially dissociates in water:

\[
\ce{HA <=> H+ + A^-}
\]

Here:

• \(\mathrm{HA}\) is the weak acid.
• \(\mathrm{A^-}\) is its conjugate base.
• \(\mathrm{H^+}\) is the hydrogen ion released during dissociation.
• \(K_a\) is the acid dissociation constant at the specified temperature.

The equilibrium expression is:

\[
K_a=\frac{[H^+][A^-]}{[HA]}
\]

Rearranging:

\[
[H^+]=K_a\frac{[HA]}{[A^-]}
\]

Taking the negative logarithm of both sides gives the Henderson-Hasselbalch equation:

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

where:

• \(pH\) measures the acidity of the solution.
• \(pK_a=-\log_{10}(K_a)\).
• \([A^-]\) is the equilibrium concentration of the conjugate base, in mol L\(^{-1}\).
• \([HA]\) is the equilibrium concentration of the weak acid, in mol L\(^{-1}\).

The equation tells you something useful: the pH depends on the ratio of conjugate base to weak acid, not simply their total concentration.

For example, if the conjugate base and weak acid have equal concentrations:

\[
\frac{[A^-]}{[HA]}=1
\]

Since \(\log_{10}(1)=0\):

\[
\boxed{pH=pK_a}
\]

This is why a buffer has a pH equal to its acid’s \(pK_a\) when the two components are present in equal equilibrium concentrations.

Why the ratio matters

Imagine two solutions containing ethanoic acid and sodium ethanoate.

Both have the same ratio, so the Henderson-Hasselbalch equation predicts the same pH.

However, solution A contains ten times as much of each component. It has a much larger supply of acid and conjugate base available to respond to additions of strong acid or base.

This is the difference between buffer pH and buffer capacity. They are related, but they are not the same thing.

02Why equilibrium shifts matter

A buffer works because its two components react with added acid or base.

Consider the equilibrium:

\[
\ce{HA <=> H+ + A^-}
\]

There are two important situations.

Adding a strong acid

Suppose you add hydrochloric acid, which dissociates completely in water and releases hydrogen ions.

The additional \(\mathrm{H^+}\) shifts the equilibrium to the left, according to Le Chatelier’s principle.

The conjugate base reacts with the added hydrogen ions:

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

This means:

• The amount of \(\mathrm{A^-}\) decreases.
• The amount of \(\mathrm{HA}\) increases.
• Most of the added hydrogen ions are consumed.

The buffer has not made the acid disappear. It has converted one buffer component into the other.

Adding a strong base

Now suppose you add sodium hydroxide.

The hydroxide ions react with the weak acid:

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

This means:

• The amount of \(\mathrm{HA}\) decreases.
• The amount of \(\mathrm{A^-}\) increases.
• The hydroxide ions are consumed.

In both cases, the buffer responds by changing its composition.

The important limitation: both components have finite amounts

Think of the conjugate base as a limited supply of hydrogen-ion catchers.

Each mole of \(\mathrm{A^-}\) can react with one mole of added \(\mathrm{H^+}\). Once the available \(\mathrm{A^-}\) has been consumed, the buffer cannot continue neutralising added acid in the same way.

The weak acid has a similar limitation when you add a strong base.

This is why you cannot simply assume the Henderson-Hasselbalch equation will work regardless of how much acid or base is added.

You must account for the available amounts first.

03When is the Henderson-Hasselbalch equation reliable?

There are three main conditions to check.

Condition 1: Both members of the conjugate pair must be present

You need an appreciable amount of both the weak acid and its conjugate base.

A solution containing only a weak acid is not a buffer of this type. Likewise, if all the conjugate base has been consumed by added strong acid, the original buffer composition no longer exists.

For example, consider a buffer initially containing:

• 0.0200 mol of \(\mathrm{HA}\)
• 0.0150 mol of \(\mathrm{A^-}\)

If you add 0.0150 mol of HCl, the reaction consumes all the conjugate base:

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

After the reaction:

• \(\mathrm{HA}=0.0350\) mol
• \(\mathrm{A^-}=0\) mol

You cannot use the Henderson-Hasselbalch equation with these amounts because the conjugate base has been exhausted.

If you add more HCl than this, excess strong acid remains in solution.

Condition 2: Neither component should be nearly exhausted

A common practical guideline is that the ratio of conjugate base to weak acid should be between 0.1 and 10:

\[
0.1\leq\frac{[A^-]}{[HA]}\leq10
\]

This corresponds approximately to:

\[
pK_a-1\leq pH\leq pK_a+1
\]

Within this range, neither component is present in an extremely small proportion relative to the other.

This is a useful rule of thumb, not an absolute mathematical boundary. A ratio outside this range does not automatically make the equation meaningless, but it should make you question whether the usual approximation is reliable.

Condition 3: The equilibrium shift must be small relative to the available amounts

This is the condition students often overlook.

The Henderson-Hasselbalch equation is derived using equilibrium concentrations. However, when solving typical buffer problems, we usually calculate the amounts immediately after the strong acid or base has reacted, then use those amounts in the equation.

We are assuming that the subsequent equilibrium shift is small enough that these amounts barely change.

Consider two situations:

Situation A: Plenty of both components

A solution contains 0.100 mol of \(\mathrm{HA}\) and 0.100 mol of \(\mathrm{A^-}\).

A small equilibrium shift will have very little effect on either amount.

Using the post-reaction amounts in the Henderson-Hasselbalch equation is a good approximation.

Situation B: Very little conjugate base remains

A solution contains 0.0190 mol of \(\mathrm{HA}\) and only 0.00100 mol of \(\mathrm{A^-}\).

Even a small additional dissociation of the weak acid could noticeably change the amount of conjugate base.

The assumption that the post-reaction amounts are approximately equal to the equilibrium amounts becomes less reliable.

This is the central limitation of the equation’s common shortcut.

A practical comparison

04Always neutralise strong acid or base before using the equation

This is probably the most useful calculation habit to develop.

When a strong acid or base is added to a buffer, do not immediately substitute the original concentrations into Henderson-Hasselbalch.

First, calculate the moles involved in the neutralisation reaction.

For an added strong acid:

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

For an added strong base:

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

Update the amounts of both buffer components, then decide whether the equation is still appropriate.

A useful bookkeeping table is:

For added NaOH, the changes are reversed.

Remember that the mole ratio can be used instead of the concentration ratio because both buffer components occupy the same final solution volume:

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

This is particularly useful when the addition changes the total solution volume.

05What happens when the approximation fails?

There is a subtle point worth understanding.

The Henderson-Hasselbalch equation does not suddenly stop being mathematically true when a buffer becomes depleted.

The problem is that the concentrations you substitute into it may no longer represent the equilibrium concentrations.

If the equilibrium shift is significant, return to the original acid dissociation expression:

\[
K_a=\frac{[H^+][A^-]}{[HA]}
\]

Set up an equilibrium calculation using the amounts remaining after the initial neutralisation reaction.

This allows you to account for the additional dissociation of the weak acid.

The distinction is:

• Henderson-Hasselbalch with post-neutralisation amounts: an approximation.
• The \(K_a\) expression with equilibrium concentrations: the more complete equilibrium calculation.

For most ordinary buffer calculations, the approximation is very useful. The important skill is knowing when its assumptions are no longer reasonable.

06Questions and solutions

Question 1

A buffer contains 0.0250 mol of a weak acid, \(\mathrm{HA}\), and 0.0400 mol of its conjugate base, \(\mathrm{A^-}\).

The acid dissociation constant is \(1.80\times10^{-5}\) at 25°C.

Calculate the pH of the buffer using the Henderson-Hasselbalch equation.

Solution 1

The pH is 4.95.

Both the weak acid and its conjugate base are present, so the Henderson-Hasselbalch equation is appropriate.

First, calculate the \(pK_a\):

\[
pK_a=-\log_{10}(1.80\times10^{-5})
\]

\[
pK_a=4.745
\]

Next, calculate the ratio:

\[
\frac{[A^-]}{[HA]}
=
\frac{0.0400}{0.0250}
=1.60
\]

Substitute:

\[
\begin{aligned}
pH&=4.745+\log_{10}(1.60)\\
&=4.745+0.204\\
&=4.949
\end{aligned}
\]

Therefore:

\[
\boxed{pH=4.95}
\]

The conjugate base is present at a higher concentration than the weak acid, so the pH is greater than the \(pK_a\).

Question 2

An ethanoic acid/ethanoate buffer initially contains:

• 0.0400 mol of ethanoic acid
• 0.0300 mol of ethanoate ions

The initial solution volume is 250.0 mL.

A student adds 20.0 mL of 0.500 mol L\(^{-1}\) sodium hydroxide.

Given \(K_a=1.80\times10^{-5}\) at 25°C:

(a) Calculate the pH after the sodium hydroxide has reacted.

(b) Explain whether the pH increases or decreases compared with its original value.

Solution 2

(a) The final pH is 4.87. (b) The pH increases because hydroxide ions consume ethanoic acid and produce ethanoate ions.

Step 1: Calculate the amount of sodium hydroxide added.

\[
\begin{aligned}
n(\ce{NaOH})&=cV\\
&=0.500\times0.0200\\
&=0.0100\text{ mol}
\end{aligned}
\]

Step 2: Write the neutralisation reaction.

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

The hydroxide ions consume ethanoic acid.

Step 3: Calculate the new amounts.

Ethanoic acid remaining:

\[
0.0400-0.0100=0.0300\text{ mol}
\]

Ethanoate produced:

\[
0.0300+0.0100=0.0400\text{ mol}
\]

Step 4: Calculate \(pK_a\).

\[
pK_a=-\log_{10}(1.80\times10^{-5})
\]

\[
pK_a=4.745
\]

Step 5: Calculate the new pH.

\[
\begin{aligned}
pH&=4.745+\log_{10}
\left(\frac{0.0400}{0.0300}\right)\\
&=4.745+\log_{10}(1.333)\\
&=4.745+0.125\\
&=4.870
\end{aligned}
\]

Therefore:

\[
\boxed{pH=4.87}
\]

(b) The pH increases because the hydroxide ions convert ethanoic acid into ethanoate ions.

The original ratio was:

\[
\frac{0.0300}{0.0400}=0.750
\]

The new ratio is:

\[
\frac{0.0400}{0.0300}=1.333
\]

The increased conjugate base-to-acid ratio produces a higher pH.

Notice that the final solution volume is 270.0 mL. It does not need to be included explicitly because both concentrations have the same final volume, which cancels in the ratio.

Question 3

A buffer initially contains:

• 0.0100 mol of ethanoic acid
• 0.00400 mol of ethanoate ions

The initial volume is 100.0 mL.

A student adds 10.0 mL of 0.600 mol L\(^{-1}\) hydrochloric acid.

Given \(K_a=1.80\times10^{-5}\) at 25°C:

(a) Determine whether the Henderson-Hasselbalch equation can be used after the addition.

(b) Calculate the approximate final pH.

Assume the excess strong acid dominates the hydrogen ion concentration.

Solution 3

(a) The Henderson-Hasselbalch equation cannot be used as an ordinary buffer approximation because all the ethanoate ions are consumed. (b) The final pH is approximately 1.74.

Step 1: Calculate the amount of hydrochloric acid added.

\[
\begin{aligned}
n(\ce{HCl})&=0.600\times0.0100\\
&=0.00600\text{ mol}
\end{aligned}
\]

Step 2: Compare the available amounts.

The solution initially contains 0.00400 mol of ethanoate ions.

The added acid supplies 0.00600 mol of hydrogen ions.

Since:

\[
0.00600>0.00400
\]

there is insufficient conjugate base to consume all the added acid.

Step 3: Calculate the excess hydrogen ions.

\[
n(\mathrm{H}^{+})_{\text{excess}}
=
0.00600-0.00400
\]

\[
=0.00200\text{ mol}
\]

The ethanoate has been completely consumed.

The remaining ethanoic acid is:

\[
0.0100+0.00400=0.0140\text{ mol}
\]

Step 4: Calculate the final solution volume.

\[
V=100.0+10.0=110.0\text{ mL}
\]

\[
V=0.1100\text{ L}
\]

Step 5: Calculate the excess hydrogen ion concentration.

\[
\begin{aligned}
[H^+]&=\frac{n}{V}\\
&=\frac{0.00200}{0.1100}\\
&=0.0182\text{ mol L}^{-1}
\end{aligned}
\]

Step 6: Calculate the pH.

\[
\begin{aligned}
pH&=-\log_{10}[H^+]\\
&=-\log_{10}(0.0182)\\
&=1.74
\end{aligned}
\]

Therefore:

\[
\boxed{pH\approx1.74}
\]

The weak acid can still dissociate slightly, but its contribution to the hydrogen ion concentration is negligible compared with the excess strong acid under these conditions.

The important point is that the buffer has reached its limit. Continuing to use the original buffer equation would ignore the excess hydrogen ions that now control the pH.

Question 4

A buffer contains 0.0200 mol of a weak acid and 0.0200 mol of its conjugate base in 200.0 mL of solution.

The entire solution is diluted with water until its volume reaches 2.00 L.

The acid dissociation constant is \(1.80\times10^{-5}\) at 25°C.

(a) Calculate the approximate pH before and after dilution.

(b) Has the total sample lost its ability to neutralise added strong acid?

(c) Compare the amount of strong acid that can be neutralised by a 100.0 mL aliquot before and after dilution.

Explain your reasoning.

Solution 4

(a) The pH remains approximately 4.74. (b) The total sample retains the same total amount of conjugate base. (c) A 100.0 mL aliquot of the diluted solution can neutralise one-tenth as much strong acid as a 100.0 mL aliquot of the original solution.

(a) Calculate the pH before dilution.

First, calculate the \(pK_a\):

\[
pK_a=-\log_{10}(1.80\times10^{-5})
\]

\[
pK_a=4.745
\]

The acid and conjugate base have equal amounts:

\[
\frac{[A^-]}{[HA]}=\frac{0.0200}{0.0200}=1
\]

Therefore:

\[
pH=4.745+\log_{10}(1)
\]

\[
\boxed{pH=4.74}
\]

Calculate the pH after dilution.

After dilution, the concentrations are:

\[
[HA]=\frac{0.0200}{2.00}
=0.0100\text{ mol L}^{-1}
\]

\[
[A^-]=\frac{0.0200}{2.00}
=0.0100\text{ mol L}^{-1}
\]

The ratio is still:

\[
\frac{[A^-]}{[HA]}=1
\]

Therefore:

\[
\boxed{pH=4.74}
\]

The pH remains approximately unchanged because both buffer components are diluted by the same factor.

The concentrations are still sufficiently large for the Henderson-Hasselbalch approximation to be reasonable here.

(b) Consider the entire sample.

The original solution contained 0.0200 mol of conjugate base.

After dilution, the entire 2.00 L sample still contains 0.0200 mol of conjugate base.

Therefore, the total amount of hydrochloric acid that can be consumed by the conjugate base before it is exhausted remains:

\[
\boxed{0.0200\text{ mol HCl}}
\]

Dilution has not removed any of the buffer components. It has simply spread them through a larger volume.

(c) Compare equal-volume aliquots.

Before dilution, a 100.0 mL aliquot contains:

\[
n(A^-)=0.0200\times\frac{100.0}{200.0}
\]

\[
=0.0100\text{ mol}
\]

After dilution, a 100.0 mL aliquot contains:

\[
n(A^-)=0.0200\times\frac{100.0}{2000}
\]

\[
=0.00100\text{ mol}
\]

The diluted aliquot contains one-tenth as much conjugate base.

It can therefore neutralise one-tenth as much strong acid before its conjugate base is exhausted.

This demonstrates an important distinction:

• The pH of a buffer depends primarily on the ratio of its components.
• Its capacity per unit volume depends on how much of those components is present in that volume.
• Diluting the entire sample does not remove its total supply of acid or conjugate base, but it reduces the capacity of an equal-volume aliquot.

For a related comparison, see How to Compare Buffers with the Same pH but Different Concentrations.

Question 5

Two buffers, S and L, contain the same weak acid and conjugate base pair.

The acid dissociation constant is:

\[
K_a=1.80\times10^{-5}
\]

Both solutions initially have equal amounts of weak acid and conjugate base.

Their compositions are:

A student adds 0.00900 mol of HCl to each buffer. Both final solution volumes are 1.00 L.

Assume ideal dilute-solution behaviour and that the strong acid reacts completely with the conjugate base before the weak-acid equilibrium is considered.

(a) Explain why both buffers initially have approximately the same pH.

(b) Calculate the amounts of weak acid and conjugate base remaining in each buffer immediately after the strong acid has reacted.

(c) Calculate the pH of buffer L using the Henderson-Hasselbalch equation. Explain why the approximation is reliable.

(d) A student applies the Henderson-Hasselbalch equation to buffer S and obtains a pH. Test whether this result is reliable by calculating the additional equilibrium dissociation of the weak acid.

Calculate the more accurate pH of buffer S.

Solution 5

Buffer L can be calculated reliably using Henderson-Hasselbalch, but buffer S requires a more complete equilibrium calculation. The more accurate final pH of buffer S is approximately 3.57.

The interesting part is that both buffers initially have the same pH, but they do not respond equally to the same acid addition.

(a) Compare their initial pH values.

Both buffers initially have equal amounts of weak acid and conjugate base.

Therefore:

\[
\frac{[A^-]}{[HA]}=1
\]

Using Henderson-Hasselbalch:

\[
pH=pK_a+\log_{10}(1)
\]

\[
pH=pK_a
\]

Calculate:

\[
pK_a=-\log_{10}(1.80\times10^{-5})
\]

\[
\boxed{pH=4.745}
\]

Both buffers initially have approximately the same pH.

(b) Calculate the amounts after adding HCl.

The reaction is:

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

The added 0.00900 mol of hydrogen ions consumes 0.00900 mol of conjugate base.

For buffer S:

\[
n(A^-)=0.0100-0.00900
\]

\[
=0.00100\text{ mol}
\]

\[
n(HA)=0.0100+0.00900
\]

\[
=0.0190\text{ mol}
\]

For buffer L:

\[
n(A^-)=0.100-0.00900
\]

\[
=0.0910\text{ mol}
\]

\[
n(HA)=0.100+0.00900
\]

\[
=0.109\text{ mol}
\]

Both buffers still contain both components.

However, the relative amounts are now quite different.

The ratio for S is:

\[
\frac{[A^-]}{[HA]}
=
\frac{0.00100}{0.0190}
=0.0526
\]

The ratio for L is:

\[
\frac{[A^-]}{[HA]}
=
\frac{0.0910}{0.109}
=0.835
\]

Buffer S has lost 90% of its original conjugate base, whereas buffer L has lost only 9%.

This is the key difference.

(c) Calculate the pH of buffer L.

The ratio is 0.835, which is within the usual practical buffer range.

The Henderson-Hasselbalch equation gives:

\[
\begin{aligned}
pH&=4.745+\log_{10}(0.835)\\
&=4.745-0.078\\
&=4.667
\end{aligned}
\]

Therefore:

\[
\boxed{pH\approx4.67}
\]

The approximation is reliable because the equilibrium shift is very small compared with the amounts of both components.

We can estimate the additional dissociation of the weak acid using:

\[
[H^+]\approx K_a\frac{[HA]}{[A^-]}
\]

\[
[H^+]\approx
1.80\times10^{-5}
\times\frac{0.109}{0.0910}
\]

\[
[H^+]\approx2.16\times10^{-5}
\text{ mol L}^{-1}
\]

This is only about 0.024% of the remaining conjugate base concentration.

The change is negligible, so using the post-neutralisation concentrations is a very good approximation.

(d) Test the Henderson-Hasselbalch result for buffer S.

If we use the equation immediately, we obtain:

\[
\begin{aligned}
pH&=4.745+\log_{10}
\left(\frac{0.00100}{0.0190}\right)\\
&=3.466
\end{aligned}
\]

However, the ratio is below 0.1, indicating that the conjugate base is becoming scarce.

We need to check whether the weak acid dissociates enough to change the available amounts significantly.

The equilibrium is:

\[
\ce{HA <=> H+ + A^-}
\]

Immediately after neutralisation, the concentrations are:

\[
[HA]=0.0190\text{ mol L}^{-1}
\]

\[
[A^-]=0.00100\text{ mol L}^{-1}
\]

Let \(x\) represent the additional concentration of weak acid that dissociates as equilibrium is established.

The equilibrium concentrations become:

Substitute these expressions into the acid dissociation constant:

\[
K_a=\frac{[H^+][A^-]}{[HA]}
\]

\[
1.80\times10^{-5}
=
\frac{x(0.00100+x)}{0.0190-x}
\]

Rearranging:

\[
x^2+0.001018x-3.42\times10^{-7}=0
\]

Solving the quadratic and taking the positive root:

\[
x=2.66\times10^{-4}\text{ mol L}^{-1}
\]

Therefore:

\[
[H^+]=2.66\times10^{-4}
\text{ mol L}^{-1}
\]

Calculate the pH:

\[
\begin{aligned}
pH&=-\log_{10}(2.66\times10^{-4})\\
&=3.575
\end{aligned}
\]

Therefore:

\[
\boxed{pH\approx3.57}
\]

Compare the two results:

The difference is approximately 0.11 pH units.

Why did this happen?

The additional dissociation produced:

\[
x=0.000266\text{ mol L}^{-1}
\]

The conjugate base initially remaining was only:

\[
0.00100\text{ mol L}^{-1}
\]

Therefore, the additional dissociation increased the conjugate base concentration by approximately 26.6%.

That is not a negligible change.

The original calculation assumed that the post-neutralisation amounts were approximately equal to the equilibrium amounts. For buffer S, that assumption was not sufficiently accurate.

This is the central lesson of the Henderson-Hasselbalch equation: having both buffer components present is not enough. Their equilibrium concentrations must also be reasonably approximated by the amounts remaining after neutralisation.

Buffer L satisfies this condition. Buffer S does not.

The next useful concept is buffer capacity: how much strong acid or base a solution can absorb before one of its buffer components becomes scarce. Understanding that distinction allows you to predict not only a buffer’s pH, but also how effectively it can resist changes in pH.