Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right)\) has one acidic hydrogen. A \(0.10 \mathrm{M}\) solution of lactic acid has a \(\mathrm{pH}\) of \(2.44 .\) Calculate \(K_{a}\)

To calculate the hydrogen ion concentration \([\mathrm{H}^+]\), we will use the pH value given: \(pH = -\log{[\mathrm{H}^+]}\) Now, we need to find \([\mathrm{H}^+]\) by using the given pH value: \(pH = 2.44\), so \([\mathrm{H}^+] = 10^{-2.44}\)

Write the dissociation reaction: \(\mathrm{CH_3CH(OH)COOH} \rightleftharpoons \mathrm{CH_3CH(OH)COO^-} + \mathrm{H^+ }\) Set up the ICE (initial, change, equilibrium) table for the given reaction: $$ \begin{array}{c|ccc} & \mathrm{CH_3CH(OH)COOH} & \mathrm{CH_3CH(OH)COO^-} & \mathrm{H^+} \\ \hline \text{Initial} & 0.10 & 0 & 10^{-2.44} \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 0.10 - x & x & 10^{-2.44} + x \end{array} $$

Using the ICE table, we can now write an expression for the equilibrium constant, \(K_a\): $$ K_a = \frac{ [\mathrm{CH_3CH(OH)COO^-}] [\mathrm{H^+}] } { [\mathrm{CH_3CH(OH)COOH}] } $$ Substitute the equilibrium values into the expression: $$ K_a = \frac{ (x)(10^{-2.44} + x)}{0.10 - x} $$ Since the \(pH\) is small, we can assume that \(x\) is much smaller than the initial concentration (\(0.10 \mathrm{M}\)), which allows us to simplify the expression: $$ K_a \approx \frac{(x)(10^{-2.44})}{0.10} $$ Now, we can plug in the value for the hydrogen ion concentration that we found in step 1: $$ x \approx 10^{-2.44} $$ Substitute this value back into the \(K_a\) expression: $$ K_a \approx \frac{(10^{-2.44})(10^{-2.44})}{0.10} $$ Finally, calculate \(K_a\): $$ K_a \approx 1.38 \times 10^{-4} $$ So, the acid dissociation constant for lactic acid is approximately \(1.38 \times 10^{-4}\).