Answer
a. $[OH^-] = 1 \times 10^{- 7}M$ : Neutral
b. $[OH^-] = 12.05M $ : Basic.
c. $[OH^-] = 8.333 \times 10^{- 16}M$ : Acidic.
d. $[OH^-] = 1.852 \times 10^{- 10}M$: Acidic.
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.
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 1 \times 10^{- 7} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$
$[OH^-] = 1 \times 10^{- 7}M$
b.
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 8.3 \times 10^{- 16} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 8.3 \times 10^{- 16}}$
$[OH^-] = 12.05 $
c.
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 12 * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 12 \times 10}$
$[OH^-] = 8.333 \times 10^{- 16}$
d.
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 5.4 \times 10^{- 5} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 5.4 \times 10^{- 5}}$
$[OH^-] = 1.852 \times 10^{- 10}$
