Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 43

Answer

$[OBr^-] \approx [H_3O^+ \approx 3.354 \times 10^{- 5} M$ $[HOBr] \approx 0.45M$

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. -$[H_3O^+] = [OBr^-] = x$ -$[HOBr] = [HOBr]_{initial} - x = 0.45 - x$ For approximation, we consider: $[HOBr] = 0.45M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$ $Ka = 2.5 \times 10^{- 9}= \frac{x * x}{ 0.45}$ $Ka = 2.5 \times 10^{- 9}= \frac{x^2}{ 0.45}$ $ 1.125 \times 10^{- 9} = x^2$ $x = 3.354 \times 10^{- 5}$ Percent ionization: $\frac{ 3.354 \times 10^{- 5}}{ 0.45} \times 100\% = 0.007454\%$ %ionization < 5% : Right approximation. Therefore: $[OBr^-] = [H_3O^+] = x = 3.354 \times 10^{- 5} $ $[HOBr] \approx 0.45M$
Small 1531332111
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.