Question

\[0.1M{\text{ }}HCl\] solution is diluted by 100 times. The pH of the solution is:
(A) 3
(B) 2
(C) 4
(D) 6

Answer 1

Hint: The acid strength of the solution is expressed in terms of pH. pH of the solution is the negative logarithm to the base 10 of the hydronium ion concentration.
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]\]
When a solution is diluted, you can obtain the concentration of the dilute solution by using the following formula.
\[{M_2} = \dfrac{{{M_1} \times {V_1}}}{{{V_2}}}\]

Complete step by step answer:
\[{\text{HCl}}\]is hydrochloric acid. When \[{\text{HCl}}\] is added to water, solution is acidic with pH being less than 7 and \[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] > {10^{ - 7}}M\]
Use the formula\[{M_2} = \dfrac{{{M_1} \times {V_1}}}{{{V_2}}}\] to calculate the molarity of diluted solution. This will give hydronium ion concentration obtained from the ionization of hydrochloric acid.
In the formula, \[{M_1},{V_1}\] refer to initial molarity and initial volume.
Similarly, \[{M_2},{V_2}\] refer to final molarity and final volume.
\[{M_2} = \dfrac{{{M_1} \times {V_1}}}{{{V_2}}} \\
= {M_1} \times \dfrac{{{V_1}}}{{{V_2}}} \\
= 0.1M \times \dfrac{1}{{100}} \\
= 0.001M \]
Here, \[\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{1}{{100}}\] as the solution is diluted 100 times.
 \[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = [{\text{HCl]}} = 0.001M\]
Calculate the pH
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = - {\log _{10}}0.001M = 3\]
Thus, the pH of the solution is 3.

Hence, the correct option is option A ).

Note: Do not consider the hydronium ion concentration from autoionization of water. This is because the hydronium ion concentration obtained from ionization of hydrochloric acid is much greater than the hydronium ion concentration obtained from autoionization of water.