Answer
- $HNO_3$ $6.0M$: $[OH^-] = 1.667 \times 10^{- 15}M$
- $NaOH$ $6.0M$: $[H_3O^+] = 1.667 \times 10^{- 15}M$
Work Step by Step
- $HNO_3$ is a strong acid, therefore: $[H_3O^+] = [HNO_3] = 6.0M$
Use the "Kw" expression to find the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 6.0 * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 6.0}$
$[OH^-] = 1.667 \times 10^{- 15}M$
- $NaOH$ is a strong base, therefore: $[OH^-] = [NaOH] = 6.0M$
Use the "Kw" expression to calculate the hydronium concentration:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 6.0 * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 6.0}$
$[H_3O^+] = 1.667 \times 10^{- 15}M$
