# Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 39

$Ka = 5.465\times 10^{- 6}$

#### 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. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x = 3.5 \times 10^{- 5} - x$ 2. Write the percent ionization equation, and find 'x': %ionization = $\frac{x}{[Initial Acid]} \times 100$ 0.85= $\frac{x}{ 0.075} \times 100$ 0.0085= $\frac{x}{ 0.075}$ $6.375\times 10^{- 4}= x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{( 6.375\times 10^{- 4})^2}{ 0.075- 6.375\times 10^{- 4}}$ $Ka = \frac{ 4.064\times 10^{- 7}}{ 0.07436}$ $Ka = 5.465\times 10^{- 6}$

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