# Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 46

$pOH = 2.216$ $pH = 11.784$

#### 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. -$[OH^-] = [N{H_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x = 2.05 - x$ For approximation, we consider: $[NH_3] = 2.05M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 2.05}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 2.05}$ $3.69 \times 10^{- 5} = x^2$ $x = 6.075 \times 10^{- 3}$ Percent ionization: $\frac{ 6.075 \times 10^{- 3}}{ 2.05} \times 100\% = 0.2963\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = x = 6.075 \times 10^{- 3}M$ 3. Calculate the pOH and pH: $pOH = -log[OH^-]$ $pOH = -log( 6.075 \times 10^{- 3})$ $pOH = 2.216$ $pH + pOH = 14$ $pH + 2.216 = 14$ $pH = 11.784$

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