Answer
(a)
$[H_3O^+] = 3.981 \times 10^{- 8}M$
$[OH^-] = 2.512 \times 10^{- 7}M$
(b)
$[H_3O^+] = 1.585 \times 10^{- 3}M$
$[OH^-] = 6.31 \times 10^{- 12}M$
Work Step by Step
(a)
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 7.4}$
$[H_3O^+] = 3.981 \times 10^{- 8}$
2. Calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 3.981 \times 10^{- 8} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 3.981 \times 10^{- 8}}$
$[OH^-] = 2.512 \times 10^{- 7}$
(b)
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.8}$
$[H_3O^+] = 1.585 \times 10^{- 3}$
2. Calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 1.585 \times 10^{- 3} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 1.585 \times 10^{- 3}}$
$[OH^-] = 6.31 \times 10^{- 12}$
