Chapter 16 - Acids and Bases - Example 16-3 - Calculating Ion Concentrations in an Aqueous Solution of a Strong Acid - Page 707: Practice Example A

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

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