Question

The pH of ${10^{ - 5}}M{\text{ HCl}}$ solution if 1 ml of it is diluted to 1000 ml is:
A ) 5
B ) 8
C ) 7.02
D ) 6.96

Answer 1

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.

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.

= \dfrac{{{{10}^{ - 5}} \times 1}}{{1000}} \\\ = {10^{ - 8}}M$$ Since the value of the hydronium ion concentration obtained from hydrochloric acid is very small, you should also consider hydronium ion concentration obtained from the autoionization of water. To the above value, add hydronium ion concentration obtained from autoionization of water. $$\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = {10^{ - 7}}M + {10^{ - 8}}M$$ Calculate the pH $${\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = - {\log _{10}}\left( {{{10}^{ - 7}}M + {{10}^{ - 8}}M} \right) = 6.96$$ Hence, the only correct option is option D ). **Thus, the pH of the solution is 6.96.** **Additional Information:** When an acid solution is diluted, the hydronium ion concentration for the diluted solution is obtained by using the formula $${M_2} = \dfrac{{{M_1} \times {V_1}}}{{{V_2}}}$$. **Note:** Do not ignore the hydronium ion concentration from autoionization of water. This is because the hydronium ion concentration obtained from ionization of hydrochloric acid is much smaller than the hydronium ion concentration obtained from autoionization of water.