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".
$Ka = \frac{[H_3O^+][OI^-]}{[HOI]}$
$2.3 \times 10^{-11} = \frac{0.045 * [OI^-]}{0.527}$
$\frac{2.3 \times 10^{-11} * 0.527}{0.045} = [OI^-]$
$[OI^-] = 2.694 \times 10^{-10}M$
