Question

How do you calculate the pH and pOH of a solution which has a hydroxide concentration of 0.0053 M?

Answer 1

Hint: The pH of a solution is its potential or power of hydrogen or hydronium ion concentration in any solution. While the pOH value is the measure of hydroxide ion concentration in a solution. The pH value ranges between $0 - 14$.

Formula used:
1. \[pH+pOH=14\] and
2. $pOH=-\log ([O{{H}^{-}}])$

Complete answer:
The pH of any solution is determined by taking the negative logarithm of its hydrogen or hydronium cation concentration, similarly pOH is determined by the negative logarithm of hydroxide anion concentration in any solution.
We have been given the hydroxide ion concentration, through which the pOH can be calculated. After that the pH can be calculated using the relationship between pH and pOH.
Given concentration of hydroxide ion, $[O{{H}^{-}}]=0.0053\,M$
Since, $pOH=-\log ([O{{H}^{-}}])$
$pOH=-\log (0.0053\,M)$
$pOH=2.28$
Now, putting this value of pOH into the relation between pH and pOH we will find the pH.
\[pH+pOH=14\]
\[pH+2.28=14\]
\[pH=14-2.28\]
\[pH=11.72\]
This shows that we are dealing with a basic solution as mentioned in the question as it contains more hydroxide anion than hydronium cations.
Hence, the pH value of the solution is 11.72 and pOH value is 2.28.

Note:
An alternate method can also be used to determine the pH value that uses ionization constant.
\[[{{H}_{3}}{{O}^{+}}].[O{{H}^{-}}]={{10}^{-14}}\] Where ${{10}^{-14}}={{K}_{w}}$ (ionization constant of water)
So the pH value will be calculated as taking out the concentration of hydronium cation and calculating its negative logarithm as:
$pH=-\log ([{{H}_{3}}{{O}^{+}}])$
Then using the relationship between pH and pOH to calculate pOH.