Question

100 ml of a solution of HCl with pH value 3 is diluted with 400 ml of water. The new pH of the solution is?
(A) 3.7
(B) 5.3
(C) 4.2
(D) 5.6

Answer 1

Hint: The pH of an acidic solution is lesser than 7.As we dilute a solution by adding water , it becomes more and more like pure water and the pH moves closer to the pH of pure water, pH 7. Thus the pH of an acidic solution such as HCl will increase on dilution and acidity will decrease.

Complete step by step solution:
- As we know, the pH scale measures how alkaline or acidic a substance is and the scale runs from 0 to 14, with a pH lower than 7 being acidic, a pH higher than 7 being alkaline and a pH of 7 being neutral. Also, the pH scale is logarithmic which means that each whole pH value below seven will be ten times more acidic than the next higher value.
- The notion of $pH$ can be described mathematically as the potential hydrogen ion concentration. That is, $pH$ is the negative logarithm of $\left[ {{H}^{+}} \right]$ and we can write this relation as follows
\[pH=-\log \left[ {{H}^{+}} \right]\]
In the question the HCl has a pH value of 3. Hence, we can modify the above relation as follows
\[-\log \left[ {{H}^{+}} \right]=3\]
 The hydrogen ion concentration in HCl can be written as
\[\left[ {{H}^{+}} \right]=1{{0}^{-3}}\]
From the concept of dilution we can write ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ where ${{M}_{1}}$ is the concentration before dilution , ${{V}_{1}}$ is the volume before dilution and ${{M}_{2}}$ is the concentration after dilution , ${{V}_{2}}$is the volume after dilution.
In the question it’s given that ${{V}_{1}}$ is 100 ml, ${{V}_{2}}$ is 500 ml. We got the value of ${{M}_{1}}$ as${{10}^{-3}}$. We need to find ${{M}_{2}}$.Lets substitute the given values in the above equation.
\[{{10}^{-3}}\times 100={{M}_{2}}\times 500\]
\[{{M}_{2}}=2\times {{10}^{-4}}\]
${{M}_{2}}$ is the hydrogen ion concentration after dilution and hence we can write as follows
\[pH=-\log \left[ {{H}^{+}} \right]=-\log \left( 2\times {{10}^{-4}} \right)=3.7\]
Therefore the new pH of the solution is 3.7

Thus the correct answer is option (A) 3.7.

Note: Keep in mind that when an alkali is diluted, the concentration of hydroxyl ions will decrease and as a result the pH of the alkaline solution decreases towards 7 making the solution less alkaline. Also, the pH of an alkaline solution cannot become lower than 7, since the water we are adding to dilute is not acidic.