What Happens to Conjugate Acid-Base Pairs When a Reaction Reverses?

Suppose ethanoic acid transfers a proton to ammonia, forming ethanoate and ammonium ions. Now reverse the reaction. Does ethanoate stop being the conjugate base of ethanoic acid and become an acid itself?

No. Ethanoate remains the conjugate base of ethanoic acid. In the reverse reaction, it acts as a base by accepting a proton from ammonium, reforming ethanoic acid.

The conjugate acid-base pairs stay the same when a reaction is reversed. What changes is the direction of proton transfer.

This distinction becomes especially important when you start comparing acid strength, equilibrium constants, and the actual direction in which a reaction proceeds.

01What happens to the conjugate pairs?

Consider the reaction between ethanoic acid and ammonia:

\[ \ce{CH3COOH + NH3 <=> CH3COO^- + NH4+} \]

In the forward direction:

• Ethanoic acid, \(\ce{CH3COOH}\), donates a proton. It acts as the acid.
• Ammonia, \(\ce{NH3}\), accepts that proton. It acts as the base.
• Ethanoate, \(\mathrm{CH_{3}COO}^{-}\), is the conjugate base of ethanoic acid.
• Ammonium, \(\ce{NH4+}\), is the conjugate acid of ammonia.

There are two conjugate pairs:

Notice that the second pair is written in the opposite order to the original reactants. This is because ammonium is the acid, while ammonia is its conjugate base.

Now reverse the reaction

Write the equation in the opposite direction:

\[ \ce{CH3COO^- + NH4+ <=> CH3COOH + NH3} \]

Follow the proton again.

• Ammonium donates a proton, forming ammonia.
• Ethanoate accepts that proton, forming ethanoic acid.

The roles in this particular proton transfer are now:

• \(\ce{NH4+}\) acts as the acid.
• \(\mathrm{CH_{3}COO}^{-}\) acts as the base.

The products are the original acid and base.

The important point is that neither conjugate pair changes its identity.

Think of each pair as two forms of the same acid-base system. One form has an extra proton compared with the other.

The direction of the reaction determines which species donates or accepts the proton. It does not change which species are conjugates.

If you need to refresh the underlying identification method, see Conjugate Acid-Base Pairs and Proton Transfer.

02Does reversing the reaction change its equilibrium?

Not automatically.

Writing a chemical equation backwards is not the same as physically making a reaction run backwards. The actual direction depends on the composition of the reaction mixture.

Consider the general proton-transfer reaction:

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

Here:

• \(\ce{HA}\) is the original acid.
• \(\ce{B}\) is the original base.
• \(\mathrm{A}^{-}\) is the conjugate base of \(\ce{HA}\).
• \(\ce{HB+}\) is the conjugate acid of \(\ce{B}\).

The forward reaction transfers a proton from \(\ce{HA}\) to \(\ce{B}\).

The reverse reaction transfers a proton from \(\ce{HB+}\) to \(\mathrm{A}^{-}\).

The equilibrium constant changes direction

For the forward reaction, the equilibrium constant is:

\[ K_{\mathrm{forward}}= \frac{[\mathrm{A}^{-}][\ce{HB+}]} {[\ce{HA}][\ce{B}]} \]

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

When you reverse the equation, the products and reactants exchange positions. The equilibrium expression therefore becomes:

\[ K_{\mathrm{reverse}}= \frac{[\ce{HA}][\ce{B}]} {[\mathrm{A}^{-}][\ce{HB+}]} \]

This gives a very useful relationship:

\[ \boxed{K_{\mathrm{reverse}}=\frac{1}{K_{\mathrm{forward}}}} \]

For example, if the forward reaction has an equilibrium constant of 100, the reverse reaction has an equilibrium constant of 0.010.

The equilibrium constant is reciprocal because the concentration ratio has been inverted.

However, the chemical species themselves have not changed their identities.

What does this tell us about the reaction?

A larger equilibrium constant means the equilibrium position favours the products of the equation as written.

It does not mean the reaction must always proceed in that direction from every possible starting mixture.

That distinction is crucial.

Acid strength determines which direction is favoured

There is another useful relationship between the equilibrium constant and acid strength.

For the general reaction:

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

the equilibrium constant can be calculated using the acid dissociation constants of the two acids:

\[ K_{\mathrm{forward}}= \frac{K_a(\ce{HA})}{K_a(\ce{HB+})} \]

Here, \(K_a\) measures the tendency of an acid to donate a proton to water.

A larger \(K_a\) indicates a stronger acid.

Therefore:

• If \(\ce{HA}\) is stronger than \(\ce{HB+}\), the forward reaction is favoured.
• If \(\ce{HB+}\) is stronger than \(\ce{HA}\), the reverse reaction is favoured.

Why?

The forward reaction transfers a proton from \(\ce{HA}\) to \(\ce{B}\). It is favoured when the resulting acid, \(\ce{HB+}\), is weaker than the original acid.

In other words, proton transfer generally favours the formation of the weaker acid and weaker conjugate base.

This is the connection between conjugate pairs and acid strength.

03Worked examples: Reversing acid-base reactions

Let’s apply these relationships to actual equilibrium calculations.

04Strength, amount, and concentration are different

When working with reversed acid-base reactions, three ideas are easy to mix up.

They describe different properties of a chemical system.

Acid strength

Acid strength is related to \(K_a\).

A strong acid has a greater tendency to donate a proton than a weak acid under the same conditions.

Its conjugate base is correspondingly weak.

Reversing the reaction does not make a weak acid strong or a strong acid weak.

At a fixed temperature and in the same solvent, the \(K_a\) of a particular acid remains unchanged.

Amount

Amount refers to the number of moles present.

For example, a container might initially contain:

• \(0.10\) mol of ethanoic acid.
• \(0.10\) mol of ammonia.
• \(0.50\) mol of ethanoate.
• \(0.50\) mol of ammonium.

Simply rewriting the equation backwards does not alter any of these amounts.

If the mixture then undergoes a net reverse reaction, the amounts change as ethanoate and ammonium are consumed.

Concentration

Concentration describes how much of a substance is present in a particular volume.

It is calculated using:

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

where:

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

Concentration matters because it determines the value of the reaction quotient, \(Q\).

Changing the concentration of one species can change the reaction quotient and cause the system to move towards equilibrium.

However, changing concentration does not automatically change the acid’s intrinsic strength.

For a more detailed comparison of these ideas, see How to Explain Acid Strength Using Equilibrium Rather Than pH Alone.

05A reliable way to analyse a reversed acid-base reaction

When an HSC question asks you what happens when an acid-base reaction is reversed, work through three separate decisions.

Decision 1: Identify the conjugate pairs.

Find the species that differ by exactly one proton.

For example:

\[ \ce{HA/A^-} \qquad \ce{HB+/B} \]

The pair identities do not change when the reaction is reversed.

Decision 2: Identify the proton donor and acceptor.

For the reverse reaction:

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

• \(\ce{HB+}\) donates the proton.
• \(\mathrm{A}^{-}\) accepts the proton.
• \(\ce{HA}\) and \(\ce{B}\) are reformed.

Decision 3: Determine the net reaction direction.

Use the reaction quotient and equilibrium constant.

If the equation itself is reversed, remember that both quantities are reciprocated:

\[ K_{\mathrm{reverse}}=\frac{1}{K_{\mathrm{forward}}} \]

\[ Q_{\mathrm{reverse}}=\frac{1}{Q_{\mathrm{forward}}} \]

These relationships ensure that both ways of writing the equation predict the same physical direction.

The following questions test these ideas in increasingly demanding situations.

06Questions and solutions

Question 1

Consider the following forward proton-transfer reaction:

\[ \ce{HCOOH + NH3 <=> HCOO^- + NH4+} \]

Formic acid, \(\ce{HCOOH}\), reacts with ammonia.

a) Write the reaction in reverse.

b) Identify the acid and base in the reverse reaction.

c) Identify both conjugate acid-base pairs.

Solution 1

The reverse reaction is:

\[ \ce{HCOO^- + NH4+ <=> HCOOH + NH3} \]

Ammonium is the acid, and formate is the base in the reverse reaction.

Ammonium donates a proton to form ammonia, while formate accepts a proton to reform formic acid.

The two conjugate pairs are:

• \(\ce{HCOOH/HCOO^-}\): formic acid and its conjugate base, formate.
• \(\ce{NH4+/NH3}\): ammonium and its conjugate base, ammonia.

The pairs remain unchanged because each pair still consists of two species differing by exactly one proton.

Question 2

Consider the general proton-transfer reaction:

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

At a particular temperature, its forward equilibrium constant is:

\[ K_{\mathrm{forward}}=4.0\times10^3 \]

a) Calculate the equilibrium constant for the reverse reaction.

b) State which side is favoured when the reverse equation is written.

c) Explain why reversing the equation does not change the identity of \(\mathrm{A}^{-}\).

Solution 2

The reverse equilibrium constant is \(2.5\times10^{-4}\), and the reverse equation favours its reactants.

Step 1: Write the relationship between the two equilibrium constants.

\[ K_{\mathrm{reverse}}= \frac{1}{K_{\mathrm{forward}}} \]

Step 2: Substitute the given value.

\[ K_{\mathrm{reverse}}= \frac{1}{4.0\times10^3} \]

\[ K_{\mathrm{reverse}}=2.5\times10^{-4} \]

Step 3: Interpret the result.

Since \(K_{\mathrm{reverse}}<1\), the equilibrium favours the reactants of the reverse equation:

\[ \ce{A^- + HB+} \]

This is consistent with the forward reaction favouring its products.

Importantly, \(\mathrm{A}^{-}\) remains the conjugate base of \(\ce{HA}\). It accepts a proton from \(\ce{HB+}\) to reform \(\ce{HA}\).

The smaller reverse equilibrium constant describes the equilibrium position, not a change in the chemical identity or intrinsic strength of \(\mathrm{A}^{-}\).

Question 3

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

\[ \ce{CH3COOH + NH3 <=> CH3COO^- + NH4+} \]

The forward equilibrium constant is:

\[ K_{\mathrm{forward}}=3.2\times10^4 \]

A \(500.0\ \mathrm{mL}\) solution contains the following amounts:

a) Calculate the concentration of each species.

b) Calculate the forward reaction quotient, \(Q_{\mathrm{forward}}\).

c) Determine the net direction of the reaction.

d) Reverse the equation and demonstrate that using the reverse reaction quotient gives the same physical prediction.

Solution 3

The net reaction proceeds forwards, towards ethanoate and ammonium. Both the forward and reverse calculations predict this same direction.

Step 1: Convert the volume into litres.

\[ V=500.0\ \mathrm{mL}=0.5000\ \mathrm{L} \]

Step 2: Calculate each concentration.

Using:

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

For ethanoic acid:

\[ [\ce{CH3COOH}]= \frac{0.0100}{0.5000} =0.0200\ \mathrm{mol\,L^{-1}} \]

For ammonia:

\[ [\ce{NH3}]= \frac{0.00500}{0.5000} =0.0100\ \mathrm{mol\,L^{-1}} \]

For ethanoate:

\[ [\mathrm{CH_{3}COO}^{-}]= \frac{0.200}{0.5000} =0.400\ \mathrm{mol\,L^{-1}} \]

For ammonium:

\[ [\ce{NH4+}]= \frac{0.200}{0.5000} =0.400\ \mathrm{mol\,L^{-1}} \]

Step 3: Calculate the forward reaction quotient.

\[ Q_{\mathrm{forward}}= \frac{[\mathrm{CH_{3}COO}^{-}][\ce{NH4+}]} {[\ce{CH3COOH}][\ce{NH3}]} \]

Substitute:

\[ Q_{\mathrm{forward}}= \frac{(0.400)(0.400)} {(0.0200)(0.0100)} \]

\[ Q_{\mathrm{forward}}= \frac{0.160}{0.000200} \]

\[ Q_{\mathrm{forward}}=800 \]

Step 4: Compare \(Q\) with \(K\).

\[ Q_{\mathrm{forward}}=800 \]

\[ K_{\mathrm{forward}}=3.2\times10^4 \]

Therefore:

\[ Q_{\mathrm{forward}}<K_{\mathrm{forward}} \]

The net reaction proceeds forwards.

Ethanoic acid and ammonia are consumed, while ethanoate and ammonium are formed.

Step 5: Check the reverse reaction.

The reverse equation is:

\[ \ce{CH3COO^- + NH4+ <=> CH3COOH + NH3} \]

Its equilibrium constant is:

\[ K_{\mathrm{reverse}}= \frac{1}{3.2\times10^4} \]

\[ K_{\mathrm{reverse}}=3.125\times10^{-5} \]

The reverse reaction quotient is:

\[ Q_{\mathrm{reverse}}= \frac{(0.0200)(0.0100)} {(0.400)(0.400)} \]

\[ Q_{\mathrm{reverse}}=0.00125 \]

Therefore:

\[ Q_{\mathrm{reverse}}>K_{\mathrm{reverse}} \]

For the reverse equation, the system moves towards its reactants, \(\mathrm{CH_{3}COO}^{-}\) and \(\ce{NH4+}\).

These are the products of the original forward equation.

Both methods give the same prediction.

This is an important consistency check: changing the way an equation is written cannot change what the actual mixture does.

Question 4

A reversible reaction is at dynamic equilibrium:

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

A student writes the equation backwards:

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

They then add a catalyst to the reaction vessel.

The student claims:

“The reverse equilibrium constant is less than one, so the reaction must now move backwards. The catalyst will make this happen faster.”

Evaluate this explanation.

Solution 4

The student’s explanation is incorrect. There is no net shift in reaction direction because the system was already at equilibrium, and adding a catalyst does not change the equilibrium position.

There are two separate mistakes.

First mistake: Assuming that rewriting the equation changes the reaction mixture.

Before the equation is reversed, the system is already at equilibrium.

This means:

\[ Q_{\mathrm{forward}}=K_{\mathrm{forward}} \]

When the equation is reversed, both quantities become their reciprocals:

\[ Q_{\mathrm{reverse}}= \frac{1}{Q_{\mathrm{forward}}} \]

\[ K_{\mathrm{reverse}}= \frac{1}{K_{\mathrm{forward}}} \]

Therefore:

\[ Q_{\mathrm{reverse}}=K_{\mathrm{reverse}} \]

The reverse equation is also at equilibrium.

Nothing has physically changed just because the equation was rewritten.

Second mistake: Assuming a catalyst changes the equilibrium position.

A catalyst provides an alternative reaction pathway with a lower activation energy.

It increases the rates of both the forward and reverse reactions.

Since the system was already at equilibrium, the forward and reverse rates remain equal.

Consequently:

• The equilibrium composition does not change.
• The equilibrium constant does not change.
• There is no net reaction in either direction.

A catalyst helps a system reach equilibrium faster, but it does not determine which side is favoured.

Question 5

Consider a hypothetical aqueous proton-transfer system:

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

At a constant temperature, its forward equilibrium constant is:

\[ K_{\mathrm{forward}}=256 \]

An equilibrium mixture is initially contained in a \(1.00\ \mathrm{L}\) vessel.

Its composition is:

The following changes are then made:

1. Water is added until the volume doubles to \(2.00\ \mathrm{L}\).
2. After the dilution, an additional \(0.0200\ \mathrm{mol}\) of \(\ce{B}\) is added. Assume the volume remains \(2.00\ \mathrm{L}\).

Assume an ideal dilute aqueous solution, negligible volume change when adding \(\ce{B}\), and constant temperature.

a) Verify that the initial mixture is at equilibrium.

b) Determine whether doubling the volume causes a net reaction. Explain why counting the number of dissolved species on each side matters.

c) Calculate the new reaction quotient after adding \(\ce{B}\).

d) Predict the subsequent net reaction direction and explain why the change in concentration does not mean that the acid strength has changed.

Solution 5

Dilution causes no net shift in this system. Adding \(\ce{B}\) afterwards causes the reaction to proceed forwards.

This question tests three connected ideas: equilibrium, dilution, and the difference between concentration and acid strength.

Step 1: Verify the initial equilibrium.

The initial volume is \(1.00\ \mathrm{L}\), so the initial concentrations are:

\[ [\ce{HA}]=0.0100\ \mathrm{M} \]

\[ [\ce{B}]=0.0100\ \mathrm{M} \]

\[ [\mathrm{A}^{-}]=0.160\ \mathrm{M} \]

\[ [\ce{HB+}]=0.160\ \mathrm{M} \]

Calculate the reaction quotient:

\[ Q= \frac{(0.160)(0.160)} {(0.0100)(0.0100)} \]

\[ Q=\frac{0.0256}{0.000100} \]

\[ Q=256 \]

Since:

\[ Q=K \]

the initial mixture is at equilibrium.

Step 2: Determine what happens when the volume doubles.

The volume increases from \(1.00\ \mathrm{L}\) to \(2.00\ \mathrm{L}\).

All four concentrations are halved because the amounts of each species initially remain unchanged.

The new concentrations are:

Step 3: Calculate the reaction quotient after dilution.

\[ Q= \frac{(0.0800)(0.0800)} {(0.00500)(0.00500)} \]

\[ Q=\frac{0.00640}{0.0000250} \]

\[ Q=256 \]

Therefore:

\[ Q=K \]

There is no net reaction.

This result is particularly interesting because students often assume dilution must favour the side with more dissolved particles.

However, count the species in the balanced equation:

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

There are two dissolved reactant species and two dissolved product species.

Dilution halves every concentration, so both the numerator and denominator of the reaction quotient are multiplied by the same factor:

\[ Q_{\mathrm{new}}= \frac{(1/2)^2[\mathrm{A}^{-}][\ce{HB+}]} {(1/2)^2[\ce{HA}][\ce{B}]} \]

The factors cancel.

This is why dilution does not shift this equilibrium under the stated ideal-solution assumptions.

Step 4: Calculate the concentration after adding \(\ce{B}\).

An additional \(0.0200\ \mathrm{mol}\) of \(\ce{B}\) is added to the \(2.00\ \mathrm{L}\) solution.

The increase in concentration is:

\[ \Delta[\ce{B}]= \frac{0.0200}{2.00} \]

\[ \Delta[\ce{B}]=0.0100\ \mathrm{M} \]

The new concentration of \(\ce{B}\) is:

\[ [\ce{B}]_{\mathrm{new}}= 0.00500+0.0100 \]

\[ [\ce{B}]_{\mathrm{new}}=0.0150\ \mathrm{M} \]

The other three concentrations remain unchanged immediately after the addition.

Step 5: Calculate the new reaction quotient.

\[ Q= \frac{(0.0800)(0.0800)} {(0.00500)(0.0150)} \]

\[ Q=\frac{0.00640}{0.0000750} \]

\[ Q=85.3 \]

Compare this with the equilibrium constant:

\[ Q=85.3 \]

\[ K=256 \]

Therefore:

\[ Q<K \]

The net reaction proceeds forwards.

Step 6: Explain the chemical changes.

The forward reaction consumes \(\ce{HA}\) and \(\ce{B}\), producing \(\mathrm{A}^{-}\) and \(\ce{HB+}\).

The system moves towards equilibrium by using up some of the added base.

Notice that adding more \(\ce{B}\) changed its concentration and the reaction quotient. It did not change the intrinsic acid strength of \(\ce{HA}\).

The equilibrium constant also remained unchanged because the temperature was constant.

The key distinction is that concentration determines the current reaction quotient, while acid strength is described by \(K_a\).

When analysing more complicated acid-base equilibria, keep these ideas separate: conjugate pairs identify the proton relationship, \(K_a\) compares acid strength, and \(Q\) compared with \(K\) predicts the net reaction direction. This is the foundation for explaining how acid-base equilibrium responds to changes in concentration, dilution, and the composition of a reaction mixture.