Answer
a)$[OH^-] = 2.941 \times 10^{- 6}M$
b)$[H_3O^+] = 1 \times 10^{- 12}M$
c)$[H_3O^+] = 1 \times 10^{- 4}M$
d)$[OH^-] = 1 \times 10^{- 7}M$
e)$[OH^-] = 1.163 \times 10^{- 10}M$
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) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 3.4 \times 10^{- 9} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 3.4 \times 10^{- 9}}$
$[OH^-] = 2.941 \times 10^{- 6}$
- Basic
b) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 2}}$
$[H_3O^+] = 1 \times 10^{- 12}$
- Basic
c) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1 \times 10^{- 10} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 10}}$
$[H_3O^+] = 1 \times 10^{- 4}$
- Acidic
d) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 1 \times 10^{- 7} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$
$[OH^-] = 1 \times 10^{- 7}$
- Neutral
e) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 8.6 \times 10^{- 5} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 8.6 \times 10^{- 5}}$
$[OH^-] = 1.163 \times 10^{- 10}$
- Acidic
