## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 16 - Acids and Bases - Exercises - Weak Acids, Weak Bases, and pH - Page 739: 22

#### Answer

$[H^+] = 1.73 \times 10^{- 12}M$ $pH = 11.76$

#### Work Step by Step

1. Using the ICE table, we got: $[OH^-] and [C_2H_5NH_3^+] = x$ and $[C_2H_5NH_2] = 0.085 - x$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = 4.3 \times 10^{- 4}= \frac{x * x}{ 8.5 \times 10^{- 2}- x}$ $Kb = 4.3 \times 10^{- 4}= \frac{x^2}{ 8.5 \times 10^{- 2}- x}$ $3.65 \times 10^{- 5} - 4.3 \times 10^{- 4}x = x^2$ $3.65 \times 10^{- 5} - 4.3 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 4.3 \times 10^{- 4})^2 - 4 * (-1) *( 3.65 \times 10^{- 5})$ $\Delta = 1.84 \times 10^{- 7} + 1.46 \times 10^{- 4} = 1.46 \times 10^{- 4}$ $x_1 = \frac{ - (- 4.29 \times 10^{- 4})+ \sqrt { 1.46 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 4.29 \times 10^{- 4})- \sqrt { 1.46 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 6.26 \times 10^{- 3} (Negative)$ $x_2 = 5.83 \times 10^{- 3}$ Therefore: $[OH^-] = 5.83 \times 10^{- 3}$ 3.Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 5.83 \times 10^{- 3})$ $pOH = 2.234$ $pH + pOH = 14$ $pH + 2.234 = 14$ $pH = 11.76$ 4. Now, find the $[H^+]$ $[H^+] = 10^{-pH}$ $[H^+] = 10^{- 11.76}$ $[H^+] = 1.73 \times 10^{- 12}$

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