## Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

# Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 51

#### Answer

$[OI^-] = 2.694 \times 10^{-10}M$

#### Work Step by Step

1. Write the Ka equation, and calculate $[OI^-]$: $Ka (HOI) = 2.3 \times 10^{-11}$ * Check appendix "F". $Ka = \frac{[H_3O^+][OI^-]}{[HOI]}$ $2.3 \times 10^{-11} = \frac{0.045 * [OI^-]}{0.527}$ $\frac{2.3 \times 10^{-11} * 0.527}{0.045} = [OI^-]$ $[OI^-] = 2.694 \times 10^{-10}M$

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