Chapter 20 The Properties of Acids and Bases - Problems - Page 770: 18

$Ka = 1.4\times 10^{- 5}$

Work Step by Step

1. Calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.78}$ $[H_3O^+] = 1.66 \times 10^{- 3}M$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CH_3(CH_2)_4COO^-] = x$ -$[CH_3(CH_2)_4COOH] = [CH_3(CH_2)_4COOH]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][CH_3(CH_2)_4COO^-]}{ [CH_3(CH_2)_4COOH]}$ $Ka = \frac{x^2}{[InitialCH_3(CH_2)_4COOH] - x}$ $Ka = \frac{( 1.66\times 10^{- 3})^2}{ 0.2- 1.66\times 10^{- 3}}$ $Ka = \frac{ 2.754\times 10^{- 6}}{ 0.1983}$ $Ka = 1.4\times 10^{- 5}$ -------

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