## Chemistry (4th Edition)

(a) $[HBr] = 0.758M$ (b) $[HBr] = 3.46 \times 10^{- 3}M$ (c) $[HBr] = 5.37 \times 10^{- 7}M$
- Since HBr is a strong acid: $[H_3O^+] = [HBr]$ Therefore, if we find $[H_3O^+]$, we get the HBr concentration. (a) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 0.12}$ $[H_3O^+] = 0.758M$ $[HBr] = 0.758M$ (b) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.46}$ $[H_3O^+] = 3.46 \times 10^{- 3}$ $[HBr] = 3.46 \times 10^{- 3}$ (c) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 6.27}$ $[H_3O^+] = 5.37 \times 10^{- 7}$ $[HBr] = 5.37 \times 10^{- 7}$