## Chemistry 10th Edition

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