Q. How to calculate the acid dissociation constant \(K_a\).

Definition. For a monoprotic weak acid HA that dissociates in water into H+ and A-, the acid dissociation constant Ka is defined by the equilibrium concentrations as

\[
\mathrm{Ka} \;=\; \frac{[\mathrm{H^+}]\,[\mathrm{A^-}]}{[\mathrm{HA}]}
\]

Step 1. Identify what information you have. Typical information sets are

– the initial (analytical) concentration of the acid, which we call \(C\), and the measured pH of the solution; or

– the initial concentration \(C\) and the equilibrium concentration(s) of H+, A-, or HA; or

– Ka given and you want pH (reverse problem).

Step 2. If you have measured pH, compute the hydronium concentration. Convert pH to \([\mathrm{H^+}]\) by

\[
[\mathrm{H^+}] \;=\; 10^{-\mathrm{pH}} .
\]

Step 3. Relate the equilibrium concentrations. For a simple monoprotic acid with no other acid or base present, every mole of A- produced corresponds to one mole of H+ produced (neglecting water autoionization when the acid is the dominant source of H+). Thus at equilibrium

\[
[\mathrm{A^-}] \;=\; [\mathrm{H^+}] ,
\qquad
[\mathrm{HA}] \;=\; C – [\mathrm{A^-}] .
\]

Step 4. Insert equilibrium concentrations into the Ka expression and compute Ka. Using the pH-based route this gives

\[
\mathrm{Ka} \;=\; \frac{[\mathrm{H^+}]\,[\mathrm{A^-}]}{[\mathrm{HA}]}
\;=\; \frac{[\mathrm{H^+}]^2}{C – [\mathrm{H^+}]} .
\]

Step 5. Practical approximation. If the acid is sufficiently weak and only a small fraction dissociates, then \([\mathrm{H^+}] \) is much smaller than \(C\). In that case approximate \(C – [\mathrm{H^+}] \approx C\) and

\[
\mathrm{Ka} \;\approx\; \frac{[\mathrm{H^+}]^2}{C} .
\]

Step 6. Algebraic determination from initial concentration (ICE table method). If you do not have pH but know the initial concentration \(C\) and want the equilibrium dissociation \(x = [\mathrm{H^+}]\), set up an ICE table conceptually:

\[
\begin{array}{lcl}
\text{initial:} & [\mathrm{HA}] = C , & [\mathrm{H^+}] = 0 , [\mathrm{A^-}] = 0 \\
\text{change:} & [\mathrm{HA}] = C – x , & [\mathrm{H^+}] = x , [\mathrm{A^-}] = x \\
\text{equilibrium:} & [\mathrm{HA}] = C – x , & [\mathrm{H^+}] = x , [\mathrm{A^-}] = x \\
\end{array}
\\
\]

Insert these into the Ka expression to get

\[
\mathrm{Ka} \;=\; \frac{x^2}{C – x} .
\]

Rearrange to obtain a quadratic equation for \(x\):

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

Solve the quadratic using the quadratic formula and choose the positive root:

\[
x \;=\; \frac{-\mathrm{Ka} + \sqrt{\mathrm{Ka}^2 + 4\,\mathrm{Ka}\,C}}{2} .
\]

Once you have \(x = [\mathrm{H^+}]\) compute pH by \( \mathrm{pH} = -\log_{10}(x)\).

Step 7. Worked numerical example (calculate Ka from measured pH). Suppose the initial acid concentration is \(C = 0.100\ \mathrm{M}\) and the measured pH is 2.87. Then

\[
[\mathrm{H^+}] \;=\; 10^{-2.87} \;\approx\; 1.35\times 10^{-3}\ \mathrm{M} .
\]

\[
[\mathrm{A^-}] \;=\; 1.35\times 10^{-3}\ \mathrm{M} ,
\qquad
[\mathrm{HA}] \;=\; 0.100 – 1.35\times 10^{-3} \;\approx\; 0.09865\ \mathrm{M} .
\]

\[
\mathrm{Ka} \;=\; \frac{(1.35\times 10^{-3})^2}{0.09865}
\;\approx\; 1.84\times 10^{-5} .
\]

Summary. The fundamental formula is

\[
\mathrm{Ka} \;=\; \frac{[\mathrm{H^+}]\,[\mathrm{A^-}]}{[\mathrm{HA}]} .
\]

To calculate Ka, determine the equilibrium concentrations (from pH and the initial concentration or by solving the ICE quadratic), substitute them into the formula, and evaluate. If dissociation is small relative to the initial concentration, use the approximation \(\mathrm{Ka}\approx [\mathrm{H^+}]^2 / C\) to simplify calculations.