# Chapter 15 - Solutions of Acids and Bases - Questions and Exercises - Exercises - Page 675: 15.58

$Ka = 1.352\times 10^{- 4}$ $pKa = 3.869$

#### Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: ** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Lactic acid] = [Lactic acid]_{initial} - x$ 2. Calculate the $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.66}$ $[H_3O^+] = 2.188 \times 10^{- 3}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Lactic acid]}$ $Ka = \frac{x^2}{[InitialLactic acid] - x}$ $Ka = \frac{( 2.188\times 10^{- 3})^2}{ 0.0376- 2.188\times 10^{- 3}}$ $Ka = \frac{ 4.786\times 10^{- 6}}{ 0.03541}$ $Ka = 1.352\times 10^{- 4}$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.352 \times 10^{- 4})$ $pKa = 3.869$

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