Q. how to calculate pH at equivalence point

Overview. The pH at the equivalence point depends on the strengths of the acid and base in the titration. There are three standard cases: strong acid titrated with strong base, weak acid titrated with strong base, and weak base titrated with strong acid. The steps below show how to calculate pH at the equivalence point for each case, with detailed algebra and an example calculation for a weak-acid / strong-base titration.

Case 1 — strong acid titrated by strong base. At the equivalence point the moles of acid and base are stoichiometrically equal, leaving only the salt of a strong acid and a strong base in solution. The resulting solution is neutral except for temperature-dependent water autoionization. At 25 degrees Celsius the ionic product of water is \(K_{\mathrm{w}} = 1.0 \times 10^{-14}\), so the hydrogen ion concentration is \([{\mathrm{H}}_3{\mathrm{O}}^{+}] = 1.0 \times 10^{-7}\) and the pH is 7. Thus at 25 °C

\[ \text{pH} = 7.00 \quad\text{(strong acid vs strong base, approximate at 25 °C)} \]

Case 2 — weak acid titrated by strong base. At the equivalence point all of the weak acid HA has been converted to its conjugate base A−. The solution contains the anion A− at concentration \(C\) (moles of A− divided by total volume). This anion hydrolyzes with water:

\[ \mathrm{A^{-}} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{HA} + \mathrm{OH^{-}} \]

The base hydrolysis constant is

\[ K_{\mathrm{b}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \]

Set up the equilibrium. Let the initial concentration of A− at equivalence be \(C\). Let \(x\) be the equilibrium concentration of \(\mathrm{OH^{-}}\). Then the equilibrium concentrations are

\[ [\mathrm{A^{-}}] = C – x, \qquad [\mathrm{HA}] = x, \qquad [\mathrm{OH^{-}}] = x. \]

Plug into the expression for \(K_{\mathrm{b}}\):

\[ K_{\mathrm{b}} = \frac{[\mathrm{HA}][\mathrm{OH^{-}}]}{[\mathrm{A^{-}}]} \;=\; \frac{x \cdot x}{C – x} \;=\; \frac{x^{2}}{C – x}. \]

Rearrange to obtain the quadratic equation in \(x\):

\[ x^{2} + K_{\mathrm{b}}\, x – K_{\mathrm{b}}\, C = 0. \]

Solve the quadratic exactly, if desired, using the quadratic formula:

\[ x = \frac{-K_{\mathrm{b}} + \sqrt{K_{\mathrm{b}}^{2} + 4 K_{\mathrm{b}} C}}{2}. \]

In most common titrations \(K_{\mathrm{b}} \ll C\) so the approximation \(C – x \approx C\) is valid and the simpler estimate can be used:

\[ x \approx \sqrt{K_{\mathrm{b}}\, C}. \]

Here \(x \approx [\mathrm{OH^{-}}]\). Compute pOH and then pH:

\[ \text{pOH} = -\log_{10}[\mathrm{OH^{-}}] \qquad\text{and}\qquad \text{pH} = 14.00 – \text{pOH}, \]

using the appropriate value of \(K_{\mathrm{w}}\) and the temperature-dependent relationship if not at 25 °C.

Example (weak acid / strong base). Suppose you start with 50.00 mL of 0.1000 M acetic acid ( \(K_{\mathrm{a}} = 1.8 \times 10^{-5}\) ) and titrate with 0.1000 M NaOH to the equivalence point. At equivalence the moles of acetate equal the initial moles of acetic acid. The total volume at equivalence is 100.00 mL, so the concentration of acetate is

\[ C = \frac{(0.1000\ \mathrm{mol/L})(0.05000\ \mathrm{L})}{0.10000\ \mathrm{L}} = 0.05000\ \mathrm{M}. \]

Compute \(K_{\mathrm{b}} = K_{\mathrm{w}} / K_{\mathrm{a}}\). Using \(K_{\mathrm{w}} = 1.0 \times 10^{-14}\) at 25 °C,

\[ K_{\mathrm{b}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}. \]

Use the approximation \([\mathrm{OH^{-}}] \approx \sqrt{K_{\mathrm{b}} C}\):

\[ [\mathrm{OH^{-}}] \approx \sqrt{(5.56 \times 10^{-10})(0.05000)} = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}\ \mathrm{M}. \]

Calculate pOH and then pH:

\[ \text{pOH} = -\log_{10}(5.27 \times 10^{-6}) = 5.28. \]

\[ \text{pH} = 14.00 – 5.28 = 8.72. \]

This result shows the equivalence-point pH is basic for a weak acid titrated with a strong base. If the approximation were not valid, use the quadratic solution for \(x\) given earlier and then compute pH exactly.

Case 3 — weak base titrated by strong acid. This is the mirror image. At equivalence the solution contains the conjugate acid \( \mathrm{BH^{+}} \) at concentration \(C\). The conjugate acid hydrolyzes:

\[ \mathrm{BH^{+}} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{B} + \mathrm{H_{3}O^{+}}. \]

The acid dissociation constant for \(\mathrm{BH^{+}}\) is \(K_{\mathrm{a}} = K_{\mathrm{w}} / K_{\mathrm{b}}\). Let \(x = [\mathrm{H_{3}O^{+}}]\) at equilibrium. The analogous derivation gives the quadratic

\[ x^{2} + K_{\mathrm{a}}\, x – K_{\mathrm{a}}\, C = 0, \]

or the approximation

\[ [\mathrm{H_{3}O^{+}}] \approx \sqrt{K_{\mathrm{a}}\, C}. \]

Then compute

\[ \text{pH} = -\log_{10} [\mathrm{H_{3}O^{+}}]. \]

Summary of practical steps to calculate pH at equivalence point.

1. Identify the titration type (strong/strong, weak/strong, or strong/weak).

2. Find the concentration \(C\) of the conjugate species present at equivalence (moles of analyte divided by total volume at equivalence).

3. If both acid and base are strong, pH ≈ 7.00 at 25 °C. If the conjugate is basic (weak acid titrated by strong base), compute \(K_{\mathrm{b}} = K_{\mathrm{w}} / K_{\mathrm{a}}\) and solve the hydrolysis equilibrium. If the conjugate is acidic (weak base titrated by strong acid), compute \(K_{\mathrm{a}} = K_{\mathrm{w}} / K_{\mathrm{b}}\) and solve that hydrolysis.

4. Use the quadratic expression \(x^{2} + K\, x – K\, C = 0\) with \(K = K_{\mathrm{b}}\) for conjugate base hydrolysis or \(K = K_{\mathrm{a}}\) for conjugate acid hydrolysis, or use the approximation \(x \approx \sqrt{K C}\) when \(K \ll C\). Here \(x\) is \([\mathrm{OH^{-}}]\) for base hydrolysis and \([\mathrm{H_{3}O^{+}}]\) for acid hydrolysis.

5. Convert \(x\) to pOH or pH as appropriate and report the pH at the equivalence point.

Notes. Always check the validity of approximations by verifying \(x \ll C\) after calculation. Use the exact quadratic solution when needed. Remember that pH at equivalence depends on temperature through \(K_{\mathrm{w}}\).