Question

pH of a solution of NaOH is 12.0. The pH of the HCl solution of the same molarity is:
(a) 2.0
(b) 12.0
(c) 10.0
(d) 1.7

Answer 1

Hint: The sum of the negative logarithm of the concentration of hydrogen ions and hydroxyl ions is taken 14. pH is the negative logarithm of the concentration of hydrogen ions and pOH is the negative logarithm of the concentration of hydroxyl ions. For finding the concentration of the solution, the concentration of hydrogen or hydroxyl ions is calculated.

Complete step by step answer:
Given the pH of the solution of NaOH = 12.0
The pH of the solution is 12 which means that the negative logarithm of concentration of hydrogen ions in the solution is 12.
So, we can write,
$pH\text{ }=\text{ }12=~~-log[{{H}^{+}}]\text{ }=12~~~$
Therefore, $[{{H}^{+}}]={{10}^{-12}}$
and we know that the sum of the negative logarithm of concentration of hydrogen and the hydroxyl ions in the solution = 14
pH + pOH = 14
Since the value of pH is 12, we can solve the value of$pOH$,
$12+pOH=14$
$pOH=2$
So, the negative logarithm of concentration of $O{{H}^{-}}$ ions is = 2
Therefore, Concentration of [$O{{H}^{-}}$] = ${{10}^{-2}}M$
Hence the molarity of the NaOH solution is ${{10}^{-2}}M$
So, for the same concentration of the HCl = ${{10}^{-2}}M$
Means concentration of [${{H}^{+}}$] = ${{10}^{-2}}M$
Now, for calculating pH,
The pH of the solution is calculated by negative logarithm of the concentration of hydrogen ions. That means,
$pH=-\log [{{H}^{+}}]$
The concentration of [${{H}^{+}}$] = 2
Putting the value in the above equation,
$pH=-\log [{{10}^{-2}}]$
This will become,
$pH=-(-2)\log [10]$
We know that the value of log 10 is 1
Hence,
$pH=2$
So, the correct option is (a) 2.0.

Note: Don’t get confused between the concentration of hydrogen and hydroxyl ions. When the base is given, the molarity of the solution is taken on the concentration of the hydroxyl ion and when acid is given the molarity of the solution is taken by the concentration of the hydrogen ions.