Calculating pH Before and After Equivalence in Strong-Acid Titrations

Twenty millilitres of acid and twenty millilitres of base have just been mixed. Is the solution neutral?

Not necessarily. Suppose the flask contains \(20.00\ \mathrm{mL}\) of \(0.1200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid, and you add \(20.00\ \mathrm{mL}\) of \(0.0800\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. The acid supplies \(2.400\times10^{-3}\ \mathrm{mol}\) of \(\ce{H3O+}\), while the base supplies only \(1.600\times10^{-3}\ \mathrm{mol}\) of \(\ce{OH-}\).

The acid is still in excess. The resulting pH is approximately 1.70, not 7.00.

This is the key to calculating pH before and after equivalence in a strong-acid titration: calculate the reacting moles first, identify which reagent remains in excess, and only then calculate the pH.

Before equivalence, excess \(\ce{H3O+}\) determines the pH. After equivalence, excess \(\ce{OH-}\) determines it. At equivalence, neither reagent is in excess.

The examples below use a strong monoprotic acid, such as \(\ce{HCl}\), titrated with a strong monohydroxide base, such as \(\ce{NaOH}\). Both are assumed to dissociate completely, solution volumes are treated as additive, and calculations use \(25^\circ\mathrm{C}\). All numerical data are illustrative.

01Start with the reaction, not the pH

Imagine adding sodium hydroxide to hydrochloric acid.

The molecular equation is:

\[
\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}
\]

Both substances are strong electrolytes, so they dissociate effectively completely in water.

The important net ionic equation is:

\[
\ce{H3O+(aq) + OH-(aq) -> 2H2O(l)}
\]

Notice the coefficients. One mole of hydronium ions reacts with one mole of hydroxide ions.

This gives a \(1:1\) mole ratio.

You can think of the reaction as a one-for-one subtraction: every mole of hydroxide added removes one mole of hydronium. The analogy is useful for keeping track of the amounts, but remember that the remaining ions are dissolved throughout the entire solution.

Calculate the equivalence volume

The equivalence point occurs when the acid and base have reacted in exactly the required stoichiometric ratio.

For a strong monoprotic acid reacting with a strong monohydroxide base:

\[
n(\ce{H3O+})=n(\ce{OH-})
\]

The amount of a substance in moles is calculated using:

\[
n=cV
\]

where:

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

At equivalence:

\[
c_{\mathrm{acid}}V_{\mathrm{acid}}
=
c_{\mathrm{base}}V_{\mathrm{eq}}
\]

Here, \(V_{\mathrm{eq}}\) is the volume of base required to reach equivalence.

Return to our opening example.

The initial acid amount is:

\[
n(\ce{H3O+})=0.1200\times0.02000
\]

\[
=2.400\times10^{-3}\ \mathrm{mol}
\]

The base concentration is \(0.0800\ \mathrm{mol\,L^{-1}}\), so the equivalence volume is:

\[
V_{\mathrm{eq}}
=
\frac{2.400\times10^{-3}}{0.0800}
\]

\[
=0.03000\ \mathrm{L}
=30.00\ \mathrm{mL}
\]

The equivalence volume is therefore \(30.00\ \mathrm{mL}\), not \(20.00\ \mathrm{mL}\).

At \(20.00\ \mathrm{mL}\) of base added, the acid is still in excess.

This is why you must establish equivalence before choosing a pH calculation. Comparing the volumes alone tells you very little unless you also know the concentrations and reaction ratio.

If the balanced equation has a different stoichiometric ratio, use its coefficients rather than automatically assuming \(1:1\).

02Calculate pH before and after equivalence

Once you know the equivalence volume, the calculation becomes a decision about which reagent remains.

Let’s build the calculations one at a time.

Before equivalence: calculate the excess acid

Before equivalence, the added hydroxide cannot neutralise all the original acid.

The amount of hydronium remaining is:

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
n(\ce{H3O+})_{\mathrm{initial}}
–
n(\ce{OH-})_{\mathrm{added}}
\]

Once you have the excess amount, calculate its concentration.

\[
[\ce{H3O+}]
=
\frac{n(\ce{H3O+})_{\mathrm{excess}}}
{V_{\mathrm{total}}}
\]

The total volume is:

\[
V_{\mathrm{total}}
=
V_{\mathrm{acid}}+V_{\mathrm{base}}
\]

Include any additional water or solution added to the flask.

Finally:

\[
\boxed{\mathrm{pH}=-\log_{10}[\ce{H3O+}]}
\]

The important detail is that you divide by the total solution volume, not just the original acid volume.

At equivalence: why is the pH 7?

At equivalence, all the original strong acid has reacted with the added strong base.

For an HCl-NaOH titration, the solution contains water, \(\mathrm{Na}^{+}\), and \(\mathrm{Cl}^{-}\).

Neither spectator ion significantly reacts with water to change the pH under the usual HSC treatment.

At \(25^\circ\mathrm{C}\), the ionic product of water is:

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

Therefore, at neutrality:

\[
[\ce{H3O+}]=[\ce{OH-}]
=1.0\times10^{-7}\ \mathrm{mol\,L^{-1}}
\]

So:

\[
\boxed{\mathrm{pH}=7.00}
\]

This value assumes \(25^\circ\mathrm{C}\). Neutrality always means equal hydronium and hydroxide concentrations, but neutral pH is not necessarily 7 at other temperatures.

After equivalence: calculate the excess base

After equivalence, all the original acid has been consumed.

Every additional mole of hydroxide remains in excess.

The amount remaining is:

\[
n(\ce{OH-})_{\mathrm{excess}}
=
n(\ce{OH-})_{\mathrm{added}}
–
n(\ce{H3O+})_{\mathrm{initial}}
\]

Now calculate the hydroxide concentration:

\[
[\ce{OH-}]
=
\frac{n(\ce{OH-})_{\mathrm{excess}}}
{V_{\mathrm{total}}}
\]

This time, you calculate pOH first:

\[
\mathrm{pOH}=-\log_{10}[\ce{OH-}]
\]

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

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

Therefore:

\[
\boxed{\mathrm{pH}=14.00-\mathrm{pOH}}
\]

A common mistake is to calculate the hydroxide concentration using every mole of base added. Some of that hydroxide has already reacted with the acid. Only the excess hydroxide contributes to the basic pH.

A single decision rule for all three regions

There is a useful way to keep the calculations organised.

Define the signed mole difference:

\[
\Delta n
=
n(\ce{H3O+})_{\mathrm{initial}}
–
n(\ce{OH-})_{\mathrm{added}}
\]

The sign tells you which reagent remains:

• If \(\Delta n>0\), acid is in excess.
• If \(\Delta n=0\), the mixture is at equivalence.
• If \(\Delta n<0\), base is in excess.

This sign convention works for the \(1:1\) reaction considered here. For a different reaction ratio, first account for the stoichiometric coefficients.

It gives you a simple decision process:

1. Calculate the initial acid moles.
2. Calculate the added base moles.
3. Find the equivalence volume.
4. Compare the added base volume with the equivalence volume.
5. Subtract the reacting moles to find the excess.
6. Divide by the total volume.
7. Calculate pH directly for excess acid, or calculate pOH first for excess base.

One limitation is worth knowing. The excess-ion calculations assume that the remaining strong acid or base dominates the contribution from water. Extremely close to equivalence, if the calculated excess concentration approaches \(10^{-7}\ \mathrm{mol\,L^{-1}}\), water autoionisation can no longer be ignored.

For ordinary calculations where the excess concentration is comfortably larger, the method above is appropriate.

03Worked examples

04Common mistakes to avoid

One practical detail: if a burette does not start at \(0.00\ \mathrm{mL}\), the volume delivered is the final reading minus the initial reading. Do not automatically use the final burette reading as the volume added.

05Questions and solutions

Question 1

A \(20.00\ \mathrm{mL}\) sample of \(0.150\ \mathrm{mol\,L^{-1}}\) nitric acid is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) potassium hydroxide.

Calculate:

(a) The equivalence volume of potassium hydroxide.

(b) The pH after \(15.00\ \mathrm{mL}\) of potassium hydroxide has been added.

Assume a temperature of \(25^\circ\mathrm{C}\).

Solution 1

The equivalence volume is \(30.00\ \mathrm{mL}\), and the pH after \(15.00\ \mathrm{mL}\) of base has been added is \(1.37\).

The reaction is:

\[
\ce{HNO3 + KOH -> KNO3 + H2O}
\]

The mole ratio is \(1:1\).

Part (a): Equivalence volume

Calculate the initial acid amount:

\[
n(\ce{H3O+})
=
0.150\times0.02000
\]

\[
=3.000\times10^{-3}\ \mathrm{mol}
\]

At equivalence, the same amount of hydroxide is required.

\[
V_{\mathrm{eq}}
=
\frac{3.000\times10^{-3}}{0.1000}
\]

\[
\boxed{V_{\mathrm{eq}}=30.00\ \mathrm{mL}}
\]

Part (b): pH after \(15.00\ \mathrm{mL}\)

The added hydroxide amount is:

\[
n(\ce{OH-})
=
0.1000\times0.01500
\]

\[
=1.500\times10^{-3}\ \mathrm{mol}
\]

Since \(15.00\ \mathrm{mL}\) is less than the equivalence volume, acid remains in excess.

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
3.000\times10^{-3}
–
1.500\times10^{-3}
\]

\[
=1.500\times10^{-3}\ \mathrm{mol}
\]

The total volume is:

\[
V_{\mathrm{total}}=20.00+15.00
\]

\[
=35.00\ \mathrm{mL}
=0.03500\ \mathrm{L}
\]

Therefore:

\[
[\ce{H3O+}]
=
\frac{1.500\times10^{-3}}{0.03500}
\]

\[
=0.04286\ \mathrm{mol\,L^{-1}}
\]

\[
\mathrm{pH}=-\log_{10}(0.04286)
\]

\[
\boxed{\mathrm{pH}=1.37}
\]

The solution remains acidic because only half the initial acid moles have been neutralised.

Question 2

A \(40.00\ \mathrm{mL}\) sample of an unknown strong monoprotic acid is titrated with \(0.1500\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

Equivalence occurs after \(32.00\ \mathrm{mL}\) of sodium hydroxide has been added.

Calculate:

(a) The initial concentration of the acid.

(b) The pH after \(35.00\ \mathrm{mL}\) of sodium hydroxide has been added.

Solution 2

The initial acid concentration is \(0.1200\ \mathrm{mol\,L^{-1}}\), and the pH after \(35.00\ \mathrm{mL}\) of base has been added is \(11.78\).

Part (a): Initial acid concentration

At equivalence, the reacting amounts are equal:

\[
n(\ce{H3O+})=n(\ce{OH-})
\]

\[
n(\ce{H3O+})
=
0.1500\times0.03200
\]

\[
=4.800\times10^{-3}\ \mathrm{mol}
\]

The original acid volume is \(0.04000\ \mathrm{L}\).

\[
c_{\mathrm{acid}}
=
\frac{4.800\times10^{-3}}{0.04000}
\]

\[
\boxed{c_{\mathrm{acid}}=0.1200\ \mathrm{mol\,L^{-1}}}
\]

Part (b): pH after \(35.00\ \mathrm{mL}\)

The amount of hydroxide added is:

\[
n(\ce{OH-})
=
0.1500\times0.03500
\]

\[
=5.250\times10^{-3}\ \mathrm{mol}
\]

Hydroxide is in excess:

\[
n(\ce{OH-})_{\mathrm{excess}}
=
5.250\times10^{-3}
–
4.800\times10^{-3}
\]

\[
=4.500\times10^{-4}\ \mathrm{mol}
\]

The total volume is:

\[
V_{\mathrm{total}}
=
0.04000+0.03500
\]

\[
=0.07500\ \mathrm{L}
\]

Therefore:

\[
[\ce{OH-}]
=
\frac{4.500\times10^{-4}}{0.07500}
\]

\[
=0.006000\ \mathrm{mol\,L^{-1}}
\]

Calculate pOH:

\[
\mathrm{pOH}
=
-\log_{10}(0.006000)
=2.2218
\]

Then:

\[
\mathrm{pH}=14.00-2.2218
\]

\[
\boxed{\mathrm{pH}=11.78}
\]

The key is that the \(32.00\ \mathrm{mL}\) equivalence volume tells you how many moles of acid were originally present. The additional \(3.00\ \mathrm{mL}\) of base supplies the excess hydroxide.

Question 3

A student performs a titration using \(30.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

They add \(20.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

Before adding more titrant, they rinse the inside walls of the flask with \(50.00\ \mathrm{mL}\) of distilled water.

Calculate:

(a) The pH after the water has been added.

(b) The additional volume of \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide needed to reach equivalence.

(c) Explain why adding water changes the pH but does not change the additional amount of sodium hydroxide required.

Solution 3

After dilution, the pH is \(2.00\), and a further \(10.00\ \mathrm{mL}\) of sodium hydroxide is required to reach equivalence.

The important distinction is that dilution changes concentration, but it does not change the number of moles of acid or base in the flask.

Part (a): pH after adding water

First calculate the original amount of acid:

\[
n(\ce{H3O+})
=
0.1000\times0.03000
\]

\[
=3.000\times10^{-3}\ \mathrm{mol}
\]

Calculate the amount of hydroxide added:

\[
n(\ce{OH-})
=
0.1000\times0.02000
\]

\[
=2.000\times10^{-3}\ \mathrm{mol}
\]

The remaining hydronium amount is:

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
3.000\times10^{-3}
–
2.000\times10^{-3}
\]

\[
=1.000\times10^{-3}\ \mathrm{mol}
\]

Adding water does not neutralise any additional acid. The excess hydronium amount remains unchanged.

The new total volume is:

\[
V_{\mathrm{total}}
=
30.00+20.00+50.00
\]

\[
=100.00\ \mathrm{mL}
=0.10000\ \mathrm{L}
\]

Therefore:

\[
[\ce{H3O+}]
=
\frac{1.000\times10^{-3}}{0.10000}
\]

\[
=0.01000\ \mathrm{mol\,L^{-1}}
\]

\[
\mathrm{pH}
=
-\log_{10}(0.01000)
\]

\[
\boxed{\mathrm{pH}=2.00}
\]

Part (b): Additional sodium hydroxide required

The remaining acid amount is still:

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
1.000\times10^{-3}\ \mathrm{mol}
\]

The reaction ratio is \(1:1\), so the same amount of hydroxide is needed.

\[
V_{\mathrm{additional}}
=
\frac{1.000\times10^{-3}}{0.1000}
\]

\[
\boxed{V_{\mathrm{additional}}=10.00\ \mathrm{mL}}
\]

Part (c): Explain the difference

The added water increases the total volume, which decreases the hydronium concentration.

That is why the pH rises.

However, the number of moles of hydronium remaining has not changed. The same amount of hydroxide is therefore required to neutralise it.

This distinction is essential in titration calculations: concentration determines pH, while the remaining reacting amount determines how much titrant is needed to reach equivalence.

Question 4

A \(25.00\ \mathrm{mL}\) sample of an unknown strong monoprotic acid is titrated with \(0.1000\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The burette initially reads \(1.20\ \mathrm{mL}\). When the pH of the mixture is measured as 2.00, the burette reads \(16.20\ \mathrm{mL}\).

Calculate:

(a) The volume of sodium hydroxide delivered when the pH is measured.

(b) The initial concentration of the acid.

(c) The equivalence volume.

(d) The pH if the titration continues until a total of \(26.20\ \mathrm{mL}\) of sodium hydroxide has been delivered.

Assume \(25^\circ\mathrm{C}\).

Solution 4

The acid concentration is \(0.07600\ \mathrm{mol\,L^{-1}}\), the equivalence volume is \(19.00\ \mathrm{mL}\), and the final pH is \(12.15\).

This problem works backwards from a measured pH to determine the original amount of acid.

Part (a): Volume of sodium hydroxide delivered

The burette reading is not the same as the volume delivered because the initial reading was not zero.

\[
V_{\mathrm{delivered}}
=
16.20-1.20
\]

\[
\boxed{V_{\mathrm{delivered}}=15.00\ \mathrm{mL}}
\]

Part (b): Initial acid concentration

The measured pH is 2.00.

Therefore:

\[
[\ce{H3O+}]
=
10^{-2.00}
\]

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

The total solution volume at this point is:

\[
V_{\mathrm{total}}
=
25.00+15.00
\]

\[
=40.00\ \mathrm{mL}
=0.04000\ \mathrm{L}
\]

The excess hydronium amount is:

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
0.0100\times0.04000
\]

\[
=4.000\times10^{-4}\ \mathrm{mol}
\]

The amount of hydroxide already added is:

\[
n(\ce{OH-})
=
0.1000\times0.01500
\]

\[
=1.500\times10^{-3}\ \mathrm{mol}
\]

The original acid amount must account for both the hydroxide that has reacted and the hydronium still remaining.

Therefore:

\[
n(\ce{H3O+})_{\mathrm{initial}}
=
1.500\times10^{-3}
+
4.000\times10^{-4}
\]

\[
=1.900\times10^{-3}\ \mathrm{mol}
\]

Now calculate the initial acid concentration:

\[
c_{\mathrm{acid}}
=
\frac{1.900\times10^{-3}}{0.02500}
\]

\[
\boxed{c_{\mathrm{acid}}=0.07600\ \mathrm{mol\,L^{-1}}}
\]

Part (c): Equivalence volume

At equivalence, the required hydroxide amount is:

\[
n(\ce{OH-})=1.900\times10^{-3}\ \mathrm{mol}
\]

Therefore:

\[
V_{\mathrm{eq}}
=
\frac{1.900\times10^{-3}}{0.1000}
\]

\[
\boxed{V_{\mathrm{eq}}=19.00\ \mathrm{mL}}
\]

Part (d): pH after \(26.20\ \mathrm{mL}\)

The titration has passed equivalence because:

\[
26.20>19.00
\]

Hydroxide is now in excess.

The total hydroxide added is:

\[
n(\ce{OH-})_{\mathrm{added}}
=
0.1000\times0.02620
\]

\[
=2.620\times10^{-3}\ \mathrm{mol}
\]

Subtract the original acid amount:

\[
n(\ce{OH-})_{\mathrm{excess}}
=
2.620\times10^{-3}
–
1.900\times10^{-3}
\]

\[
=7.200\times10^{-4}\ \mathrm{mol}
\]

The total volume is:

\[
V_{\mathrm{total}}
=
25.00+26.20
\]

\[
=51.20\ \mathrm{mL}
=0.05120\ \mathrm{L}
\]

Calculate the hydroxide concentration:

\[
[\ce{OH-}]
=
\frac{7.200\times10^{-4}}{0.05120}
\]

\[
=0.01406\ \mathrm{mol\,L^{-1}}
\]

Calculate pOH:

\[
\mathrm{pOH}
=
-\log_{10}(0.01406)
\]

\[
=1.852
\]

Finally:

\[
\mathrm{pH}=14.00-1.852
\]

\[
\boxed{\mathrm{pH}=12.15}
\]

The important reasoning step was reconstructing the original acid amount. The measured pH gave the excess acid concentration, but the total acid initially present also included the amount already neutralised.

Question 5

A \(20.00\ \mathrm{mL}\) sample of \(0.1200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid is titrated with \(0.0800\ \mathrm{mol\,L^{-1}}\) sodium hydroxide at \(25^\circ\mathrm{C}\).

A student makes the following claim:

If I calculate the pH at a volume exactly \(1.00\ \mathrm{mL}\) before equivalence and another volume exactly \(1.00\ \mathrm{mL}\) after equivalence, the two pH values must be equally far from 7.00.

Their reasoning is that the same number of moles of acid and base will remain in excess at these two points.

Answer the following:

(a) Calculate the equivalence volume.

(b) Calculate the pH at \(29.00\ \mathrm{mL}\) of sodium hydroxide added.

(c) Calculate the pH at \(31.00\ \mathrm{mL}\) of sodium hydroxide added. Determine whether the student’s claim is correct.

(d) Find the exact volume of sodium hydroxide that must be added after equivalence for the pH to be the mirror image of the pH calculated in part (b). In other words, the two pH values must add to 14.00.

Explain why the required volume is not simply \(31.00\ \mathrm{mL}\).

Solution 5

The student’s claim is incorrect. The pH values at \(29.00\ \mathrm{mL}\) and \(31.00\ \mathrm{mL}\) are approximately 2.79 and 11.20, respectively.

The exact mirror-image point occurs after \(31.0417\ \mathrm{mL}\) of sodium hydroxide has been delivered.

The subtle issue is that equal excess mole amounts do not necessarily produce equal excess concentrations. The total solution volumes are different.

Part (a): Equivalence volume

First calculate the initial acid amount:

\[
n(\ce{H3O+})
=
0.1200\times0.02000
\]

\[
=2.400\times10^{-3}\ \mathrm{mol}
\]

The reaction ratio is \(1:1\).

Therefore:

\[
V_{\mathrm{eq}}
=
\frac{2.400\times10^{-3}}{0.0800}
\]

\[
\boxed{V_{\mathrm{eq}}=30.00\ \mathrm{mL}}
\]

Part (b): pH at \(29.00\ \mathrm{mL}\)

This point is \(1.00\ \mathrm{mL}\) before equivalence.

The hydroxide amount added is:

\[
n(\ce{OH-})
=
0.0800\times0.02900
\]

\[
=2.320\times10^{-3}\ \mathrm{mol}
\]

The remaining acid amount is:

\[
n(\ce{H3O+})_{\mathrm{excess}}
=
2.400\times10^{-3}
–
2.320\times10^{-3}
\]

\[
=8.000\times10^{-5}\ \mathrm{mol}
\]

The total volume is:

\[
V_{\mathrm{total}}
=
20.00+29.00
\]

\[
=49.00\ \mathrm{mL}
=0.04900\ \mathrm{L}
\]

Therefore:

\[
[\ce{H3O+}]
=
\frac{8.000\times10^{-5}}{0.04900}
\]

\[
=1.6327\times10^{-3}\ \mathrm{mol\,L^{-1}}
\]

The pH is:

\[
\mathrm{pH}
=
-\log_{10}(1.6327\times10^{-3})
\]

\[
\boxed{\mathrm{pH}=2.787}
\]

Part (c): pH at \(31.00\ \mathrm{mL}\)

This point is \(1.00\ \mathrm{mL}\) after equivalence.

The hydroxide amount added is:

\[
n(\ce{OH-})
=
0.0800\times0.03100
\]

\[
=2.480\times10^{-3}\ \mathrm{mol}
\]

The excess hydroxide amount is:

\[
n(\ce{OH-})_{\mathrm{excess}}
=
2.480\times10^{-3}
–
2.400\times10^{-3}
\]

\[
=8.000\times10^{-5}\ \mathrm{mol}
\]

Notice that this is exactly the same excess mole amount as in part (b).

However, the total volume is now:

\[
V_{\mathrm{total}}
=
20.00+31.00
\]

\[
=51.00\ \mathrm{mL}
=0.05100\ \mathrm{L}
\]

The hydroxide concentration is:

\[
[\ce{OH-}]
=
\frac{8.000\times10^{-5}}{0.05100}
\]

\[
=1.5686\times10^{-3}\ \mathrm{mol\,L^{-1}}
\]

Calculate pOH:

\[
\mathrm{pOH}
=
-\log_{10}(1.5686\times10^{-3})
\]

\[
=2.804
\]

Therefore:

\[
\mathrm{pH}=14.00-2.804
\]

\[
\boxed{\mathrm{pH}=11.196}
\]

The two pH values are not exact mirror images:

\[
2.787+11.196=13.983
\]

The student’s reasoning is tempting because the excess mole amounts are identical.

But the concentration of an excess species depends on both its amount and the volume in which it is dissolved.

The after-equivalence solution has a larger total volume, so its hydroxide concentration is slightly lower.

Part (d): Find the exact mirror-image point

For the two pH values to be mirror images around 7.00:

\[
\mathrm{pH}_{\mathrm{before}}
+
\mathrm{pH}_{\mathrm{after}}
=14.00
\]

Since:

\[
\mathrm{pH}_{\mathrm{before}}
=
-\log_{10}[\ce{H3O+}]
\]

and:

\[
\mathrm{pH}_{\mathrm{after}}
=
14.00+\log_{10}[\ce{OH-}]
\]

the concentrations must be equal:

\[
[\ce{H3O+}]_{\mathrm{before}}
=
[\ce{OH-}]_{\mathrm{after}}
\]

From part (b), the required concentration is:

\[
[\ce{H3O+}]
=
1.6327\times10^{-3}\ \mathrm{mol\,L^{-1}}
\]

Now let \(x\) represent the additional volume of sodium hydroxide, in mL, beyond equivalence.

The initial acid volume plus the equivalence volume is:

\[
20.00+30.00=50.00\ \mathrm{mL}
\]

Before equivalence, the excess acid amount is proportional to \(1.00\ \mathrm{mL}\) of titrant.

After equivalence, the excess hydroxide amount is proportional to \(x\ \mathrm{mL}\) of titrant.

Since the titrant concentration is the same in both cases, the required concentration equality becomes:

\[
\frac{1.00}{50.00-1.00}
=
\frac{x}{50.00+x}
\]

Rearrange:

\[
\frac{1}{49.00}
=
\frac{x}{50.00+x}
\]

Cross-multiply:

\[
50.00+x=49.00x
\]

\[
50.00=48.00x
\]

\[
x=\frac{50.00}{48.00}
\]

\[
x=1.04167\ \mathrm{mL}
\]

This is the additional volume beyond equivalence.

Therefore, the total volume delivered is:

\[
V_{\mathrm{base}}
=
30.00+1.04167
\]

\[
\boxed{V_{\mathrm{base}}=31.0417\ \mathrm{mL}}
\]

At this volume, the excess hydroxide concentration equals the hydronium concentration calculated in part (b).

Consequently:

\[
\mathrm{pOH}=2.787
\]

\[
\mathrm{pH}=14.00-2.787
\]

\[
\boxed{\mathrm{pH}=11.213}
\]

The two pH values are now exact mirror images to the precision of the calculation.

The deeper lesson is that equal distances in titrant volume do not guarantee equal distances in pH. You must account for the changing total solution volume as well as the excess mole amount.

06Where to go next

You now have a method for calculating pH at any ordinary point before or after equivalence in a strong monoprotic acid-strong base titration.

The next useful step is understanding how these individual calculations produce the full titration curve. In particular, you should be able to connect the steep pH change near equivalence to the rapid change from excess acid to excess base.

For that broader explanation, see Strong Acid-Strong Base Titration Curves for HSC Chemistry.

One important distinction remains: if the acid is weak rather than strong, the before-equivalence calculation changes because both the weak acid and its conjugate base are present. That creates a buffer region, which requires a different approach. See Weak Acid-Strong Base Titrations: HSC Chemistry Guide.