Answer
a) $[OH^-] = 4 \times 10^{- 11}M$, Acidic.
b)$[OH^-] = 5 \times 10^{- 15}M$, Acidic.
c) $[H_3O^+] = 1.786 \times 10^{- 6}M$, Acidic.
d)$[H_3O^+] = 6.667 \times 10^{- 12}M$, Basic.
e) $[H_3O^+] = 1 \times 10^{- 7}M$, Neutral.
Work Step by Step
If $[H_3O^+] > [OH^-]$, the solution is acidic.
If $[H_3O^+] < [OH^-]$, the solution is basic.
If $[H_3O^+] = [OH^-]$, the solution is neutral.
a) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 2.5 \times 10^{- 4} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 2.5 \times 10^{- 4}}$
$[OH^-] = 4 \times 10^{- 11}$
- Acidic
b) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 2.0 * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 2.0}$
$[OH^-] = 5 \times 10^{- 15}$
- Acidic
c) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 5.6 \times 10^{- 9} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 5.6 \times 10^{- 9}}$
$[H_3O^+] = 1.786 \times 10^{- 6}$
- Acidic
d) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.5 \times 10^{- 3} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.5 \times 10^{- 3}}$
$[H_3O^+] = 6.667 \times 10^{- 12}$
- Basic
e) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1 \times 10^{- 7} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$
$[H_3O^+] = 1 \times 10^{- 7}$
- Neutral
