Answer
(a) $[HBr] = 6.16 \times 10^{-5}M$
(b) $[HBr] = 2.81 \times 10^{-4}M$
(c) $[HBr] = 0.104M$
Work Step by Step
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)
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 0.98}$
$[H_3O^+] = 0.104$
$[HNO_3] = 0.104M$
