#### Answer

$[I^-] = 0.0025 \space M$
$$[OH^-] = 4.0 \times 10^{-12} \space M$$
$$pH = 2.60 $$

#### Work Step by Step

Since HI is a strong electrolyte:
$[HI] = [I^-] = [H_3O^+]$
$[I^-] = 0.0025 \space M$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 2.5 \times 10^{-3} } = 4.0 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 0.0025 ) = 2.60 $$