Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 16 - Acid-Base Equilibria - Exercises - Page 720: 16.79a

Answer

$[OH^-]= 6.325 \times 10^{- 4}M$ $pH = 10.80$

Work Step by Step

- Identify the significant electrolytes in $NaOBr$ - $Na^+$: Insignificant acid. - $BrO^-$: Weak base. Therefore, to calculate the pH, we just need to consider a $0.10M$ $BrO^-$ solution. 1. Since $BrO^-$ is the conjugate base of $HBrO$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $ 2.5\times 10^{- 9} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 2.5\times 10^{- 9}}$ $K_b = 4.0\times 10^{- 6}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [HBrO] = x$ -$[BrO^-] = [BrO^-]_{initial} - x = 0.1 - x$ For approximation, we consider: $[BrO^-] = 0.1M$ 3. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HBrO]}{ [BrO^-]}$ $Kb = 4 \times 10^{- 6}= \frac{x * x}{ 0.1}$ $Kb = 4 \times 10^{- 6}= \frac{x^2}{ 0.1}$ $ 4 \times 10^{- 7} = x^2$ $x = 6.325 \times 10^{- 4}$ Percent ionization: $\frac{ 6.325 \times 10^{- 4}}{ 0.1} \times 100\% = 0.6325\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HBrO] = x = 6.325 \times 10^{- 4}M $ $[BrO^-] \approx 0.1M$ 4. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 6.325 \times 10^{- 4})$ $pOH = 3.20$ $pH + pOH = 14$ $pH + 3.199 = 14$ $pH = 10.80$
Small 1532429381
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.