Question

On adding of the following, the pH of 20 ml of 0.1 N HCl will not alter?
A. 1 ml of 1N HCl
B. 20 ml of distilled water
C. 1 ml of 0.1N NaOH
D. 500 ml of HCl of pH=1

Answer 1

Hint: We know that normality is the number of equivalence of solute that is present in one liter of the solution. It is expressed by ‘N’. We can calculate normality by gram equivalence of solute divided by volume of solution in liters. Mathematically we can write it as follows:
     ${\text{Normality}}\,{\text{ = }}\,\dfrac{{{\text{Gram}}\,{\text{equivalence}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{solution}}\,{\text{in}}\,{\text{liters}}}}$
Unit of normality is gram equivalent per liter or equivalence per liter.

Complete step by step answer:
Given that 20 ml of 0.1 N HCl is added to a solution to form a solution of pH which is the same as the pH of 20 ml of 0.1 N HCl solution.
For the 1st option, addition of 1 ml of 1N HCl. This makes the solution more acidic thus pH will be decreased.
For the second option, 20 ml of distilled water. This makes the solution basic thus the pH of the solution will be increased.
For the third option, 1 ml of 0.1N NaOH. This makes the solution neutralization thus pH of the solution becomes equal to 7. And coming to the last option D, 500 ml of HCl of pH is equal to 1.
Now we will calculate the pH of 20 ml of 0.1 N HCl.
     $
  {\text{pH = }} - \log \left[ {{{\text{H}}^{\text{ + }}}} \right] \\
   = 1 \\
 $
Option D gives that pH value of the solution is 1. Thus this solution is added to the given solution and initial solution and resulting solution has equal pH.

Therefore the correct option is D.

Note:
When volume ${{\text{V}}_{\text{1}}}$ of a solution of normality ${{\text{N}}_{\text{1}}}$ is mixed with a volume ${{\text{V}}_2}$ of another solution of normality ${{\text{N}}_2}$ then we write the normality ${{\text{N}}_3}$ of the resulting solution is,
     ${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}{\text{ = }}{{\text{N}}_{\text{3}}}\left( {{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{V}}_{\text{2}}}} \right)$