How to Write Brønsted-Lowry Equations with Water

Add aqueous ammonia to water and the pH rises. Yet ammonia, \(\ce{NH3}\), contains no oxygen and no hydroxide ions. So where does the \(\mathrm{OH}^{-}\) come from?

Before writing an equation, predict which substance loses a proton: ammonia or water.

It is water. Ammonia accepts a proton to become \(\ce{NH4+}\), while water loses a proton to become \(\mathrm{OH}^{-}\).

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH^-(aq)} \]

This is the key to writing Brønsted-Lowry equations with water: work out which substance transfers a proton, then show what happens to the water molecule.

The direction of proton transfer determines whether water becomes hydronium, \(\ce{H3O+}\), or hydroxide, \(\mathrm{OH}^{-}\).

01The two patterns for writing equations with water

There are two common situations when a substance reacts with water.

1. An acid transfers a proton to water

An acid donates a proton, \(\mathrm{H}^{+}\). Water accepts that proton and becomes a hydronium ion.

The general equation is:

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

Here:

• \(\ce{HA}\) is the acid.
• \(\ce{H2O}\) is the Brønsted-Lowry base.
• \(\ce{H3O+}\) is the conjugate acid of water.
• \(\mathrm{A}^{-}\) is the conjugate base of the original acid.

For example, hydrochloric acid reacts with water:

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

Hydrochloric acid loses a proton and becomes \(\mathrm{Cl}^{-}\). Water gains that proton and becomes \(\ce{H3O+}\).

Notice that water is acting as a base here, even though it is not usually thought of as a basic substance.

2. A base accepts a proton from water

A base accepts a proton. Water donates one and becomes a hydroxide ion.

The general equation is:

\[ \ce{B + H2O <=> BH+ + OH^-} \]

Here:

• \(\ce{B}\) is the base.
• \(\ce{H2O}\) is the Brønsted-Lowry acid.
• \(\ce{BH+}\) is the conjugate acid of the original base.
• \(\mathrm{OH}^{-}\) is the conjugate base of water.

Ammonia follows this pattern:

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH^-(aq)} \]

Ammonia gains a proton, while water loses one.

The difference is worth remembering:

If you remember only one thing when writing these equations, remember this table.

02A reliable four-step method

When an HSC question asks you to write an acid-base equation in water, use the following method.

Step 1: Identify whether the dissolved substance is donating or accepting a proton.

Look at the substance’s chemical structure and its Brønsted-Lowry behaviour.

An acid donates \(\mathrm{H}^{+}\). A base accepts \(\mathrm{H}^{+}\).

Don’t assume that every substance containing hydrogen is an acid, or that every base must contain hydroxide.

Step 2: Write the conjugate species.

If a substance loses one proton, its charge decreases by one.

If it gains one proton, its charge increases by one.

For example:

\[ \ce{HF -> F^- + H+} \]

\[ \ce{NH3 + H+ -> NH4+} \]

These are bookkeeping representations of the proton transfer, not complete equations showing the reaction with water.

Step 3: Work out what happens to water.

This is the step students most often miss.

• If the dissolved substance is an acid, water gains \(\mathrm{H}^{+}\) and becomes \(\ce{H3O+}\).
• If the dissolved substance is a base, water loses \(\mathrm{H}^{+}\) and becomes \(\mathrm{OH}^{-}\).

Step 4: Choose the correct arrow and check the equation.

The arrow depends on how extensively the reaction proceeds.

• Strong acids ionise essentially completely in water, so use a forward arrow.
• Weak acids ionise partially, so use an equilibrium arrow.
• Weak bases react partially with water, so use an equilibrium arrow.

Finally, check that the number of each type of atom and the total charge are equal on both sides.

For a broader explanation of proton donation, acceptance, and conjugate pairs, see Brønsted-Lowry Acids and Bases: HSC Chemistry Guide.

03Worked examples

04What an experiment can tell you about the reaction

The reason these equations matter becomes clearer when you consider a simple pH investigation.

Imagine placing a calibrated pH probe into a beaker containing distilled water. The probe connects to a meter that displays the pH.

A pipette is then used to add a measured amount of aqueous ammonia. After gently stirring, the pH is measured again.

What should happen?

The prediction is that adding ammonia will increase the pH.

If ammonia reacts with water to produce hydroxide ions, the solution should become more basic.

Illustrative readings might look like this:

These are illustrative values, not experimental measurements.

The ammonia solution becomes basic, supporting the idea that hydroxide ions are being produced.

But there is a problem with interpreting the result.

A pH meter measures the acidity of the solution. It cannot tell you exactly how the hydroxide ions were formed.

Consider the comparison with sodium hydroxide.

Sodium hydroxide already contains hydroxide ions. When it dissolves, its ions separate:

\[ \ce{NaOH(s) -> Na+(aq) + OH^-(aq)} \]

Ammonia contains no hydroxide ions. Instead, it reacts with water:

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH^-(aq)} \]

Both solutions can have the same pH, but the chemical processes are different.

Estimating the hydroxide concentration

Suppose the ammonia solution has an illustrative pH of 10.50.

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

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

Therefore:

\[ \begin{aligned} \mathrm{pOH}&=14.00-10.50\\ &=3.50 \end{aligned} \]

Using the dilute-solution approximation:

\[ \begin{aligned} [\mathrm{OH}^{-}]&=10^{-3.50}\\ &=3.2\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This tells us the approximate hydroxide concentration.

It does not tell us that every ammonia molecule has reacted.

The improved investigation uses a calibrated probe, a water control, and a comparison solution. These controls help distinguish a genuine pH change from measurement drift and show why identical pH readings do not establish identical reaction mechanisms.

The conclusion comes from combining the measurement with the chemical composition of the reactants.

This is exactly why the Brønsted-Lowry equation must show water participating in the reaction.

05Strength, amount, and concentration are different

There is another important distinction when choosing the correct equation.

A concentrated acid is not necessarily a strong acid. A dilute acid is not necessarily a weak acid.

These terms describe different properties.

Strength determines the extent of reaction

A strong acid ionises essentially completely in water.

A weak acid only partially ionises.

For example:

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

Hydrochloric acid uses a forward arrow because it is strong.

Compare this with ethanoic acid:

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

Ethanoic acid is weak, so an equilibrium arrow is appropriate.

The same distinction applies to bases. A weak molecular base such as ammonia only partially reacts with water.

Concentration tells you how much is present per litre

Concentration 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.

Rearranging gives:

\[ n=cV \]

Consider two hydrochloric acid samples.

Sample A contains 50.0 mL of 0.200 mol L\(^{-1}\) HCl.

Sample B contains 200 mL of 0.0500 mol L\(^{-1}\) HCl.

For sample A:

\[ n=0.200\times0.0500 =0.0100\ \mathrm{mol} \]

For sample B:

\[ n=0.0500\times0.200 =0.0100\ \mathrm{mol} \]

Both samples contain the same amount of hydrochloric acid, but their concentrations are different.

Both are still strong acids.

Diluting hydrochloric acid does not turn it into a weak acid. It changes the concentration, not the intrinsic acid strength.

06Water can act as either an acid or a base

Water has a special property: it can both donate and accept protons.

A substance capable of doing both is called amphiprotic.

You have already seen water acting in two different ways:

As a base:

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

Water accepts a proton.

As an acid:

\[ \ce{NH3 + H2O <=> NH4+ + OH^-} \]

Water donates a proton.

This means water does not have one permanent Brønsted-Lowry role. Its role depends on the substance reacting with it.

What happens when the other substance is also amphiprotic?

Consider the dihydrogen phosphate ion, \(\mathrm{H_{2}PO_{4}}^{-}\).

It can donate a proton:

\[ \ce{H2PO4^- + H2O <=> H3O+ + HPO4^2-} \]

Here, \(\mathrm{H_{2}PO_{4}}^{-}\) acts as an acid and water acts as a base.

However, it can also accept a proton:

\[ \ce{H2PO4^- + H2O <=> H3PO4 + OH^-} \]

Here, \(\mathrm{H_{2}PO_{4}}^{-}\) acts as a base and water acts as an acid.

Both equations are balanced and chemically meaningful.

But don’t make the mistake of assuming that both reactions occur to the same extent simply because both are possible.

The relative acid and base strengths determine which pathway is more favoured.

If a question asks for both possible reactions, write both. If it asks which behaviour predominates, you need information about the relevant equilibrium constants.

07Common mistakes when writing equations with water

Before moving to practice questions, check these common errors.

One final distinction is particularly useful.

Not every substance that produces hydroxide in water reacts with water through proton transfer.

For example, sodium hydroxide already contains hydroxide ions. Its dissolution is:

\[ \ce{NaOH(s) -> Na+(aq) + OH^-(aq)} \]

Ammonia behaves differently:

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH^-(aq)} \]

The first equation shows ionic dissociation. The second shows a Brønsted-Lowry proton-transfer reaction.

For more detail on these two processes and the correct arrows for strong and weak acids, see Ionic Equations for Acid and Base Dissociation in Water.

08Questions and solutions

Question 1

Ethanoic acid, \(\ce{CH3COOH}\), is a weak acid.

Write its Brønsted-Lowry equation when dissolved in water.

Identify:

a) The Brønsted-Lowry acid and base.

b) The two conjugate acid-base pairs.

Solution 1

The equation is:

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

Ethanoic acid donates one proton from its carboxyl group, leaving the ethanoate ion, \(\mathrm{CH_{3}COO}^{-}\).

Water accepts this proton and becomes hydronium, \(\ce{H3O+}\).

Therefore:

• Acid: \(\ce{CH3COOH}\)
• Base: \(\ce{H2O}\)
• Conjugate base: \(\mathrm{CH_{3}COO}^{-}\)
• Conjugate acid: \(\ce{H3O+}\)

The conjugate pairs are:

\[ \ce{CH3COOH/CH3COO^-} \]

\[ \ce{H3O+/H2O} \]

Because ethanoic acid is weak, the equilibrium arrow is required.

Question 2

Ammonium chloride, \(\ce{NH4Cl}\), is dissolved in water.

a) Write the ionic equation for the dissolution of ammonium chloride.

b) Write the Brønsted-Lowry equation for the reaction of the ammonium ion with water.

c) Identify the acid, base, and conjugate pairs in the proton-transfer reaction.

d) Explain why the chloride ion does not appear in the proton-transfer equation.

Solution 2

The two equations describe different processes.

a) Dissolution

\[ \ce{NH4Cl(s) -> NH4+(aq) + Cl^-(aq)} \]

Ammonium chloride is an ionic compound. When it dissolves, its existing ions separate.

b) Proton transfer

The ammonium ion can donate a proton to water:

\[ \ce{NH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)} \]

c) Identify the acid and base.

The ammonium ion loses a proton, so it acts as the Brønsted-Lowry acid.

Water gains the proton, so it acts as the Brønsted-Lowry base.

The conjugate pairs are:

\[ \ce{NH4+/NH3} \]

\[ \ce{H3O+/H2O} \]

d) Explain the role of chloride.

The chloride ion does not participate in the proton transfer.

It remains unchanged in solution, so it is a spectator ion.

The key distinction is that dissolving an ionic compound and reacting one of its ions with water are separate processes. Writing them separately makes the chemistry much clearer.

Question 3

Two separate 1.00 L solutions are prepared at \(25^\circ\mathrm{C}\):

• Solution A contains 0.0100 mol L\(^{-1}\) hydrochloric acid.
• Solution B contains 1.00 mol L\(^{-1}\) ethanoic acid.

For ethanoic acid, use:

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

Assume hydrochloric acid ionises completely and use the approximation \(x\ll C\) for ethanoic acid.

a) Write the equation for each acid reacting with water, using the correct arrow.

b) Calculate the amount of acid in each solution.

c) Calculate the approximate pH of each solution.

d) Explain why the more concentrated solution does not have the lower pH.

Solution 3

Hydrochloric acid is stronger, even though the ethanoic acid solution is more concentrated and contains more moles.

a) Write the equations.

For hydrochloric acid:

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

For ethanoic acid:

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

Hydrochloric acid is strong, so it uses a forward arrow.

Ethanoic acid is weak, so it uses an equilibrium arrow.

b) Calculate the amount of each acid.

Use:

\[ n=cV \]

For hydrochloric acid:

\[ n=0.0100\times1.00 \]

\[ n=0.0100\ \mathrm{mol} \]

For ethanoic acid:

\[ n=1.00\times1.00 \]

\[ n=1.00\ \mathrm{mol} \]

Solution B contains 100 times as many moles of acid.

c) Calculate the pH values.

For hydrochloric acid, complete ionisation gives:

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

Therefore:

\[ \mathrm{pH}=-\log_{10}(0.0100)=2.00 \]

For ethanoic acid, use:

\[ [\ce{H3O+}]\approx\sqrt{K_aC} \]

Substituting:

\[ \begin{aligned} [\ce{H3O+}]&\approx\sqrt{1.8\times10^{-5}\times1.00}\\ &=4.24\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \mathrm{pH}=-\log_{10}(4.24\times10^{-3}) \]

\[ \mathrm{pH}\approx2.37 \]

d) Explain the difference.

The hydrochloric acid has a lower pH because it ionises essentially completely.

The ethanoic acid is much more concentrated, but only a small fraction of its molecules ionise.

Consequently, the dilute strong acid produces a higher hydronium concentration than the concentrated weak acid.

This demonstrates why concentration, amount, and strength must be considered separately.

Question 4

A student prepares a 0.0200 mol L\(^{-1}\) aqueous methylamine solution.

An illustrative pH measurement at \(25^\circ\mathrm{C}\) gives:

\[ \mathrm{pH}=11.44 \]

Assume:

• \(\mathrm{pH+pOH}=14.00\).
• The solution behaves ideally.
• The only significant source of hydroxide ions is the reaction of methylamine with water.

a) Write the Brønsted-Lowry equation for methylamine in water.

b) Calculate the hydroxide ion concentration.

c) Estimate the percentage of methylamine molecules that have accepted a proton.

d) Explain why a basic pH does not mean methylamine has reacted completely.

Solution 4

Approximately 13.8% of the methylamine molecules have accepted a proton. The solution is basic, but most methylamine remains unreacted.

a) Write the equation.

Methylamine accepts a proton from water:

\[ \ce{CH3NH2(aq) + H2O(l) <=> CH3NH3+(aq) + OH^-(aq)} \]

Water acts as the acid because it donates the proton.

b) Calculate the hydroxide concentration.

First, calculate pOH:

\[ \begin{aligned} \mathrm{pOH}&=14.00-11.44\\ &=2.56 \end{aligned} \]

Then:

\[ [\mathrm{OH}^{-}]=10^{-2.56} \]

\[ [\mathrm{OH}^{-}]=2.75\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

c) Calculate the percentage of methylamine that reacted.

The equation shows that each methylamine molecule accepting a proton produces one hydroxide ion.

Therefore:

\[ [\ce{CH3NH3+}]\approx[\mathrm{OH}^{-}] \]

The fraction of methylamine that reacted is:

\[ \frac{2.75\times10^{-3}}{0.0200} \]

Converting to a percentage:

\[ \begin{aligned} \%\text{ reacted} &=\frac{2.75\times10^{-3}}{0.0200}\times100\\ &=13.8\% \end{aligned} \]

d) Explain why the solution is basic without complete reaction.

The reaction produces hydroxide ions, increasing the solution’s basicity.

However, methylamine is a weak base, so the reaction only proceeds partially.

Approximately 86.2% of the original methylamine remains unprotonated under these assumptions.

A pH above 7 tells you that the solution is basic. It does not tell you that the base has reacted completely.

Question 5

A student prepares an aqueous solution containing 0.0200 mol of sodium dihydrogen phosphate, \(\ce{NaH2PO4}\), in 1.00 L of water.

The dihydrogen phosphate ion, \(\mathrm{H_{2}PO_{4}}^{-}\), is amphiprotic.

The following equilibrium constants apply at \(25^\circ\mathrm{C}\):

\[ K_{a1}(\ce{H3PO4})=7.1\times10^{-3} \]

\[ K_{a2}(\mathrm{H_{2}PO_{4}}^{-})=6.3\times10^{-8} \]

\[ K_w=1.0\times10^{-14} \]

The student makes the following argument:

“Because \(\mathrm{H_{2}PO_{4}}^{-}\) can donate and accept protons, it must react equally in both directions. Therefore, the solution should be neutral.”

a) Write the two possible Brønsted-Lowry equations involving \(\mathrm{H_{2}PO_{4}}^{-}\) and water.

b) Calculate the base ionisation constant, \(K_b\), for \(\mathrm{H_{2}PO_{4}}^{-}\) acting as a base.

c) Compare the acid and base equilibrium constants. Which behaviour is more strongly favoured?

d) Evaluate the student’s claim that the solution must be neutral.

e) The same 0.0200 mol of \(\ce{NaH2PO4}\) is now dissolved in 10.0 L of water instead of 1.00 L. Explain which properties change and which do not. Would dilution change the relative acid and base strengths of \(\mathrm{H_{2}PO_{4}}^{-}\)?

Assume ideal dilute behaviour and compare the two specified proton-transfer pathways at the same temperature.

Solution 5

The dihydrogen phosphate ion behaves more strongly as an acid than as a base. Its solution is therefore acidic rather than neutral under the stated conditions.

The important point is that being amphiprotic does not mean being equally strong as an acid and a base.

a) Write both possible equations.

First, \(\mathrm{H_{2}PO_{4}}^{-}\) can donate a proton to water:

\[ \ce{H2PO4^- + H2O <=> H3O+ + HPO4^2-} \]

Here:

• \(\mathrm{H_{2}PO_{4}}^{-}\) is the acid.
• \(\ce{H2O}\) is the base.

Alternatively, \(\mathrm{H_{2}PO_{4}}^{-}\) can accept a proton from water:

\[ \ce{H2PO4^- + H2O <=> H3PO4 + OH^-} \]

Here:

• \(\mathrm{H_{2}PO_{4}}^{-}\) is the base.
• \(\ce{H2O}\) is the acid.

Both equations are balanced.

b) Calculate \(K_b\).

When \(\mathrm{H_{2}PO_{4}}^{-}\) acts as a base, it accepts a proton and forms \(\ce{H3PO4}\).

Its conjugate acid is therefore \(\ce{H3PO4}\).

For a conjugate acid-base pair:

\[ K_aK_b=K_w \]

Therefore:

\[ K_b=\frac{K_w}{K_{a1}} \]

Substituting:

\[ \begin{aligned} K_b&=\frac{1.0\times10^{-14}}{7.1\times10^{-3}}\\ &=1.41\times10^{-12} \end{aligned} \]

This is the base ionisation constant for \(\mathrm{H_{2}PO_{4}}^{-}\).

c) Compare the acid and base strengths.

The acid ionisation constant is:

\[ K_a=6.3\times10^{-8} \]

The base ionisation constant is:

\[ K_b=1.41\times10^{-12} \]

Compare the two:

\[ \frac{K_a}{K_b} = \frac{6.3\times10^{-8}}{1.41\times10^{-12}} \]

\[ \frac{K_a}{K_b}\approx4.5\times10^4 \]

The acid equilibrium constant is approximately 45,000 times greater than the base equilibrium constant.

Therefore, the proton-donating behaviour is much more strongly favoured than the proton-accepting behaviour.

d) Evaluate the student’s claim.

The student’s conclusion is incorrect.

Amphiprotic means that a substance can both donate and accept a proton. It does not mean that the two reactions occur to the same extent.

The equilibrium constants demonstrate that \(\mathrm{H_{2}PO_{4}}^{-}\) has a much stronger tendency to donate a proton than to accept one under the stated conditions.

Consequently, the solution is acidic rather than neutral.

The mistake was treating the ability to perform two different chemical roles as evidence that both roles are equally strong.

e) Explain the effect of dilution.

The initial concentration is:

\[ C=\frac{0.0200}{1.00} =0.0200\ \mathrm{mol\,L^{-1}} \]

After dilution to 10.0 L:

\[ C=\frac{0.0200}{10.0} =0.00200\ \mathrm{mol\,L^{-1}} \]

The amount of sodium dihydrogen phosphate remains unchanged at 0.0200 mol.

However, its concentration decreases by a factor of ten.

The important distinction is that the equilibrium constants do not change simply because the solution is diluted.

At constant temperature:

\[ K_a=6.3\times10^{-8} \]

and

\[ K_b=1.41\times10^{-12} \]

remain unchanged.

Dilution changes the concentrations of the dissolved species and can change the resulting pH. It does not turn a weak acid into a strong acid, or make the substance equally strong as an acid and a base.

The acid equilibrium constant remains much larger than the base equilibrium constant.

This is the final distinction to understand: an amphiprotic substance can have two possible Brønsted-Lowry roles without having equal acid and base strengths.

Once you can write the correct equation with water, the next step is understanding how far the reaction proceeds. That means using \(K_a\), \(K_b\), concentration, and equilibrium calculations to move beyond simply writing a balanced equation and explain what actually happens in solution.