${\rm{1}}\;{\rm{cc}}$ of ${\rm{0}}{\rm{.1}}\;{\rm{N}}$ ${\rm{HCl}}$ is added to ${\rm{1}}\;{\rm{litre}}$ solution of sodium chloride. The ${\rm{pH}}$ of the resulting solution will be:
A. $7$
B. $0$
C. $10$
D. $4$
 Given:
- Volume of ${\rm{HCl}}$ added: ${{\rm{V}}_{{\rm{HCl}}}} = {\rm{1}}\;{\rm{cc}}$
- Concentration of ${\rm{HCl}}$ added: ${{\rm{N}}_{{\rm{HCl}}}} = {\rm{0}}{\rm{.1}}\;{\rm{N}}$
- Volume of ${\rm{NaCl}}$ taken: ${{\rm{V}}_{{\rm{NaCl}}}} = {\rm{1}}\;{\rm{litre}}$

Hint:
We know that the ${\rm{pH}}$ of a solution is related to the concentration of ${H^ + }$ or $O{H^ - }$ ions in the solution and that relationship can be used to do so.

Complete step by step solution:
First of all we will change the units of volume for which following conversion factor is given:
$\dfrac{{{\rm{1 }}{\rm{L}}}}{{{\rm{1000}}\;{\rm{cc}}}}$
We can use the above conversion factor to change the units of volume of ${\rm{HCl}}$ added as follows:
$\left( {\dfrac{{{\rm{1 }}{\rm{L}}}}{{{\rm{1000}}\;{\rm{cc}}}}} \right) \times {\rm{1}}\;{\rm{cc}} = 0.001{\rm{ }}{\rm{L}}$

Now, as \[0.001{\rm{ }}{\rm{L}}\] of \[HCl\] is added to \[1\;L\] of \[{\rm{NaCl}}\], we can calculate the total volume of the solution by adding the two volumes as follows:
$
{{\rm{V}}_{{\rm{total}}}} = {{\rm{V}}_{{\rm{NaCl}}}} + {{\rm{V}}_{{\rm{HCl}}}}\\
 = 1{\rm{ }}{\rm{L}} + 0.001{\rm{ }}{\rm{L}}\\
 = {\rm{1}}{\rm{.001}}\;{\rm{L}}
$

We know for \[HCl\], molarity can be taken as equal to normality itself for it is a monobasic acid. So, we can write:
\[{{\rm{M}}_{{\rm{HCl}}}} = {\rm{0}}{\rm{.1}}\;{\rm{M}}\]
Now, as we are adding a small amount of \[HCl\] solution to give a large volume, it will get diluted or we can say its concentration would change. For calculating its concentration in the resulting solution we will use the equation for dilution that can be written as follows:
${M_1}{V_1} = {M_2}{V_2}$
Here, ${M_1}$ and ${M_2}$ are the concentrations and ${V_1}$ and ${V_2}$ are the volumes before and after the dilution respectively.

We can rearrange the above equation for final concentration as follows:
${M_2} = \dfrac{{{M_1}{V_1}}}{{{V_2}}}$
Now, we can substitute the values to calculate the final concentration as follows:
$
{M_2} = \dfrac{{\left( {{\rm{0}}{\rm{.1}}\;{\rm{M}}} \right)\left( {0.001{\rm{ }}{\rm{L}}} \right)}}{{{\rm{1}}{\rm{.001}}\;{\rm{L}}}}\\
 = {\rm{9}}{\rm{.99}} \times {\rm{1}}{{\rm{0}}^{ - 5}}\;{\rm{M}}
$

We know that \[{\rm{NaCl}}\] is a neutral salt and doesn’t affect the $pH$ of the solution. So, the $pH$ of the solution would be resulting from the concentration of ${H^ + }$ ions that are coming from $HCl$. It is a strong acid so we can write:
$
\left[ {{H^ + }} \right] = \left[ {HCl} \right]\\
 = {\rm{9}}{\rm{.99}} \times {\rm{1}}{{\rm{0}}^{ - 5}}\;{\rm{M}}
$

Finally, we can use substitute the above value in the formula of $pH$ and calculate the same as follows:
$
pH = - \log \left( {\left[ {{H^ + }} \right]/M} \right)\\
 = - \log \left( {{\rm{9}}{\rm{.99}} \times {\rm{1}}{{\rm{0}}^{ - 5}}} \right)\\
 = {\rm{4}}{\rm{.0}}
$

Hence, the $pH$ of the solution is ${\rm{4}}{\rm{.0}}$ which makes option D to be the correct one.

Note:
We have to use the proper conversion factors and units while performing the calculations as $pH$ gets highly affected by a single decimal point.