Q. how to calculate half equivalence point

Definition. The half equivalence point in an acid base titration is the point at which exactly half of the initial moles of the analyte have been neutralized by the titrant. At that point the concentrations of the weak acid and its conjugate base are equal. For a weak acid titrated with a strong base the pH at the half equivalence point equals the pKa of the weak acid. For a weak base titrated with a strong acid the pOH at the half equivalence point equals the pKb of the weak base, and the pH is 14 minus pKb under standard conditions.

General step by step procedure to calculate the half equivalence point.

Step 1 Identify the type of titration and the relevant acid dissociation constant. Determine whether you are titrating a weak acid with a strong base, or a weak base with a strong acid. Obtain the acid dissociation constant \(K_{\mathrm{a}}\) for the weak acid, or the base dissociation constant \(K_{\mathrm{b}}\) for the weak base.

Step 2 Compute the volume of titrant that corresponds to the half equivalence point when volumes and concentrations are given. Let the initial analyte concentration be \(C_{\mathrm{sample}}\) and its initial volume be \(V_{\mathrm{sample}}\). The initial moles of analyte are

\[ n_{\mathrm{initial}} = C_{\mathrm{sample}} \times V_{\mathrm{sample}}. \]

At the half equivalence point the moles of titrant added equal half of the initial moles of analyte. If the titrant concentration is \(C_{\mathrm{titrant}}\), then the volume of titrant at the half equivalence point is

\[ V_{\mathrm{half}} = \frac{n_{\mathrm{initial}}}{2\,C_{\mathrm{titrant}}} = \frac{C_{\mathrm{sample}}\,V_{\mathrm{sample}}}{2\,C_{\mathrm{titrant}}}. \]

Step 3 Use the Henderson Hasselbalch relation to find pH or pOH. For a weak acid HA with conjugate base A minus the acid dissociation equilibrium is

\[ K_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}. \]

Take logarithms and rearrange to obtain the Henderson Hasselbalch equation

\[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}, \]

where \(\mathrm{p}K_{\mathrm{a}} = -\log K_{\mathrm{a}}\). At the half equivalence point the amounts and therefore the concentrations of HA and A minus are equal, so the ratio is unity and the logarithm term is zero. Thus

\[ \mathrm{pH}_{\text{half}} = \mathrm{p}K_{\mathrm{a}}. \]

For a weak base B being titrated by a strong acid the analogous result is

\[ \mathrm{pOH}_{\text{half}} = \mathrm{p}K_{\mathrm{b}}, \]

and then the pH is

\[ \mathrm{pH}_{\text{half}} = 14.00 – \mathrm{p}K_{\mathrm{b}}, \]

provided you use the customary \( \mathrm{p}K \) scale with \( \mathrm{p}K_{\mathrm{w}} = 14.00 \) at the chosen temperature. If the temperature differs significantly adjust \( \mathrm{p}K_{\mathrm{w}} \) accordingly.

Step 4 Practical calculation workflow summary.

1. Compute initial moles \( n_{\mathrm{initial}} = C_{\mathrm{sample}}\,V_{\mathrm{sample}} \). 2. Compute the titrant volume at the half equivalence point \( V_{\mathrm{half}} = \dfrac{C_{\mathrm{sample}}\,V_{\mathrm{sample}}}{2\,C_{\mathrm{titrant}}} \). 3. If the analyte is a weak acid with known \(K_{\mathrm{a}}\) compute \( \mathrm{p}K_{\mathrm{a}} = -\log K_{\mathrm{a}} \). Then \( \mathrm{pH}_{\text{half}} = \mathrm{p}K_{\mathrm{a}} \). 4. If the analyte is a weak base with known \(K_{\mathrm{b}}\) compute \( \mathrm{p}K_{\mathrm{b}} = -\log K_{\mathrm{b}} \). Then \( \mathrm{pOH}_{\text{half}} = \mathrm{p}K_{\mathrm{b}} \) and \( \mathrm{pH}_{\text{half}} = 14.00 – \mathrm{p}K_{\mathrm{b}} \).

Worked example. Suppose you titrate \(50.0\) milliliters of \(0.100\) molar acetic acid \( \mathrm{CH_{3}COOH} \) with \(0.100\) molar sodium hydroxide. The acid dissociation constant is \(K_{\mathrm{a}} = 1.8\times 10^{-5}\).

Step A compute initial moles.

\[ n_{\mathrm{initial}} = 0.100\ \mathrm{mol\,L^{-1}} \times 0.0500\ \mathrm{L} = 0.00500\ \mathrm{mol}. \]

Step B compute volume of titrant at half equivalence.

\[ V_{\mathrm{half}} = \frac{0.00500\ \mathrm{mol}}{2 \times 0.100\ \mathrm{mol\,L^{-1}}} = \frac{0.00500}{0.200}\ \mathrm{L} = 0.0250\ \mathrm{L} = 25.0\ \mathrm{mL}. \]

Step C compute pKa and pH at half equivalence.

\[ \mathrm{p}K_{\mathrm{a}} = -\log\left(1.8\times 10^{-5}\right) \approx 4.74. \]

Therefore at the half equivalence point

\[ \mathrm{pH}_{\text{half}} = \mathrm{p}K_{\mathrm{a}} \approx 4.74. \]

Notes and caveats. The equality \( \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} \) at half equivalence is exact only when activity coefficients can be neglected and when the concentrations of HA and A minus are equal. This is an excellent approximation in typical classroom and laboratory titrations at moderate ionic strength and dilute solutions. For very concentrated solutions or when activity coefficients are significant use activities instead of concentrations and adjust accordingly.