1cc of 0.1 N HCl is added to the 99cc solution of NaCl. The pH of the resulting solution will be :

Hint: We need to understand the common ion effect observed when two salts have the same ion or ions. Identify the common ion in the above aqueous solution. Write the concentration of hydrogen ions in accordance to Le chatelier's principle if the acid used is a weak acid.

Complete step-by-step answer:
The aqueous solution consists of the salt, sodium chloride and its parent acid, hydrochloric acid. It is important to know that sodium chloride does not contribute to the pH of the solution. This is because the salt is chemically neutral i.e. pH = 7.
So, the factor for deciding the pH is hydrochloric acid which consists of hydrogen ions.
We will write the dissociation of hydrochloric acid in the solution.
$\text{HCl }\to \text{ }{{\text{H}}^{-}}\text{ + C}{{\text{l}}^{-}}$
The total volume of the solution becomes 100 cc( 1cc of HCl and 99c of NaCl).
Hence the concentration of hydrogen ions becomes,
$[{{\text{H}}^{+}}]\text{ = }\dfrac{0.1}{100}\text{ M = 0}\text{.001 M}$
$\text{pH = -log(0}\text{.001)}$
So, pH = 3.

Therefore, the correct answer is option (B).

Note: 1cc denotes 1 cubic centimetre of volume occupied. 1cc is equivalent to 1mL of volume occupied as well. On the other hand, 1 ${{\text{m}}^{\text{3}}}$ of volume is equivalent to 1000L i.e. ${{10}^{3}}\text{ L}$.
It is important to know that the common ion effect is observed when the electrolyte is weak and the dissociation is incomplete due to interfering or common ions. However, the common ion effect is not applicable as HCl is a strong acid and has complete dissociation in spite of NaCl being present in the solution.