Question

Calculate the pOH of ${{10}^{-2}}$ M monobasic acid if value of ${{K}_{w}}$ at ${{90}^{\circ }}C$ is ${{10}^{-12}}$.
(a) 2
(b) 10
(c) 12
(d) 7

Answer 1

Hint: ${{K}_{w}}$is the symbol of the ionic product of water and it is equal to the sum of the concentration of hydrogen ions and concentration of hydroxyl ions in the solution. So the formula can be used as ${{K}_{w}}=pOH+pH$.

Complete step by step answer:
In the question, it is mentioned that the acid is monobasic which means that one molecule of the acid can donate one hydrogen ion or we can say that the concentration of the acid is equal to the concentration of hydrogen ions in the solution. And the concentration of the hydrogen ions in the solution is given ${{10}^{-2}}$ M. Therefore we know that the negative logarithm of the concentration of hydrogen ions in the solution is equal to the pH of the solution.
This can be written as:
$pH=-\log [{{H}^{+}}]$
Now putting the value of hydrogen ions we get:
$pH=-\log [{{10}^{-2}}]$
$pH=2\log 10$
We know that log 10 is 1, so:
$pH=2$
${{K}_{w}}$is the symbol of an ionic product of water and it is equal to the sum of concentration of hydrogen ions and concentration of hydroxyl ions in the solution.
It can be written as:
${{K}_{w}}=pOH+pH$
Now we are given the value of ionic product of water as ${{10}^{-12}}$, so this is written as:
$pH+pOH=12$
We know the value of pH is 2, so putting the value we get:
 $2+pOH=12$
$pOH=12-2$
$pOH=10$
So the value of pOH is 10.

Therefore, the correct answer is an option (b) 10.

Note: The value of ${{K}_{w}}$is ${{10}^{-12}}$, and in equation form it is converted as:
${{K}_{w}}=-\log [{{10}^{-12}}]$
${{K}_{w}}=12\log 10$
Value of log 10 is one, so
${{K}_{w}}=12$
So the equation becomes:
$pH+pOH=12$.