Answer
a. $[H_3O^+] = 6.7 \times 10^{- 15}M$: Basic.
b. $[H_3O^+] = 2.8\ M$: Acidic
c. $[H_3O^+] = 1.0 \times 10^{- 7}M$: Neutral
d. $[H_3O^+] = 1.4 \times 10^{- 11}M$: Basic
Work Step by Step
$[H_3O^+]$ represents the same thing as $[H^+]$.
- If $[H_3O^+] = [OH^-]$: Neutral
- If $[H_3O^+] > [OH^-]$: Acidic
- If $[H_3O^+] < [OH^-]$: Basic.
a.
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.5 * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.5 \times 10^{- 0}}$
$[H_3O^+] = 6.667 \times 10^{- 15}M$
b.
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 3.6 \times 10^{- 15} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 3.6 \times 10^{- 15}}$
$[H_3O^+] = 2.778M$
c.
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.0 \times 10^{- 7} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$
$[H_3O^+] = 1 \times 10^{- 7}$
d.
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 7.3 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 7.3 \times 10^{- 4}}$
$[H_3O^+] = 1.37 \times 10^{- 11}M$
