## Chemistry 10th Edition

$[HNO_3] = 3.02 \times 10^{-4}$
1. Find $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.52}$ $[H_3O^+] = 3.02 \times 10^{- 4}M$ 2. Since this is a strong acid: $[HNO_3] = [H_3O^+] = 3.02 \times 10^{-4}$