Q. How to calculate \(\mathrm{p}K_a\).

Step 1. What pKa means. The quantity \( \mathrm{p}K_{\mathrm{a}} \) is the negative base ten logarithm of the acid dissociation constant \( \mathrm{K}_{\mathrm{a}} \). By definition write the formula as a short block equation.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; -\log_{10}\!\left(\mathrm{K}_{\mathrm{a}}\right) \]

Step 2. Derivation from the equilibrium expression. A weak acid HA dissociates according to the equilibrium expression written in concentration form. Use square bracket notation inside math delimiters for concentrations.

\[ \mathrm{K}_{\mathrm{a}} \;=\; \dfrac{[H^{+}]\,[A^{-}]}{[HA]} \]

Take minus the base ten logarithm of both sides. Use properties of logarithms to separate the terms. Recognize the definitions of pKa and pH so the algebra becomes transparent.

\[ -\log_{10}\!\left(\mathrm{K}_{\mathrm{a}}\right) \;=\; -\log_{10}\!\left([H^{+}]\right) \;-\; \log_{10}\!\left(\dfrac{[A^{-}]}{[HA]}\right) \]

Replace the logarithmic terms by the standard symbols. By definition \( \mathrm{p}K_{\mathrm{a}} = -\log_{10}\!\left(\mathrm{K}_{\mathrm{a}}\right) \) and \( \mathrm{pH} = -\log_{10}\!\left([H^{+}]\right) \). Substitute those to obtain Henderson Hatchell form.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; \mathrm{pH} \;-\; \log_{10}\!\left(\dfrac{[A^{-}]}{[HA]}\right) \]

Step 3. Practical formulas you will use. There are two common ways to calculate pKa depending on the information you have.

1. If you know the acid dissociation constant then use the direct definition.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; -\log_{10}\!\left(\mathrm{K}_{\mathrm{a}}\right) \]

2. If you know the solution pH and the ratio of conjugate base to acid then use the rearranged Henderson Hatchell equation.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; \mathrm{pH} \;-\; \log_{10}\!\left(\dfrac{[A^{-}]}{[HA]}\right) \]

Step 4. Worked numeric example from a given Ka. For example: suppose \( \mathrm{K}_{\mathrm{a}} = 1.8\times 10^{-5} \). Compute the pKa step by step.

Compute the logarithm in two pieces. Use the identity \( \log_{10}\!\left(a\times 10^{b}\right) = \log_{10}\!a + b \). Here \( a = 1.8 \) and \( b = -5 \). Evaluate numeric values carefully.

\[ \log_{10}\!\left(1.8\times 10^{-5}\right) \;=\; \log_{10}\!1.8 \;+\; (-5) \;\approx\; 0.2553 \;-\; 5 \;=\; -4.7447 \]

Apply the negative sign to get pKa.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; -\left(-4.7447\right) \;\approx\; 4.7447 \]

Round as appropriate to obtain \( \mathrm{p}K_{\mathrm{a}} \approx 4.74 \). That completes the calculation from Ka.

Step 5. Worked numeric example from pH and ratio. For example: suppose the solution pH is 5.0 and the concentration ratio equals ten meaning \( [A^{-}]/[HA] = 10 \). Use the Henderson Hatchell rearrangement to find pKa.

\[ \mathrm{p}K_{\mathrm{a}} \;=\; 5.0 \;-\; \log_{10}\!10 \;=\; 5.0 \;-\; 1.0 \;=\; 4.0 \]

Step 6. Special experimental point. In an acid titration the half equivalence point gives equal amounts of acid and conjugate base so \( [A^{-}]/[HA] = 1 \). The logarithm of one is zero. Therefore at the half equivalence point the measured pH equals the pKa. This provides a common experimental method to determine pKa from a titration curve.

Summary. Use \( \mathrm{p}K_{\mathrm{a}} = -\log_{10}\!\left(\mathrm{K}_{\mathrm{a}}\right) \) when Ka is known. Use \( \mathrm{p}K_{\mathrm{a}} = \mathrm{pH} – \log_{10}\!\left([A^{-}]/[HA]\right) \) when pH and the acid base ratio are known. At the half equivalence point pH equals pKa. These formulas and the worked examples above show step by step how to calculate pKa in typical situations.