## Chemistry (4th Edition)

(a) $[HBr] = 6.16 \times 10^{-5}M$ (b) $[HBr] = 2.81 \times 10^{-4}M$ (c) $[HBr] = 0.104M$
Since $HNO_3$ is a strong acid: $[HNO_3] = [H_3O^+]$ Therefore, we need to find the hydronium ion concentration. (a) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.21}$ $[H_3O^+] = 6.16 \times 10^{- 5}$ $[HNO_3] = 6.16 \times 10^{-5}M$ (b) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.55}$ $[H_3O^+] = 2.81 \times 10^{- 4}$ $[HNO_3] = 2.81 \times 10^{-4}M$ (c) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 0.98}$ $[H_3O^+] = 0.104$ $[HNO_3] = 0.104M$