Question

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.
pH= 11.22

Answer 1

Answer: The value of $[H_{3}O^{+}]$ is $6.025 \times 10^{-12}$ M and $[OH^{-}]$ is $1 \times 10^{-14}$.

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 11.22. So, the value of concentration of hydrogen ions is calculated as follows.

$pH = - log [H^{+}]\\11.22 = - log [H^{+}]\\conc. H^{+} = 6.025 \times 10^{-12}$ M

Let the value $6.025 \times 10^{-12}$ is considered as equal to 0. Hence, the relation between pH and pOH value is as follows.

pH + pOH = 14

0 + pOH = 14

pOH = 14

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, $[OH^{-}]$ is calculated as follows.

$pOH = - log [OH^{-}]\\14 = - log [OH^{-}]\\conc. OH^{-} = 1 \times 10^{-14} M$

Thus, we can conclude that the value of $[H_{3}O^{+}]$ is $6.025 \times 10^{-12}$ M and $[OH^{-}]$ is $1 \times 10^{-14}$.