Question

1mL of 0.01N HCl is added to 999 mL solution of 0.1N $N{a_2}S{O_4}$. The pH of the resulting solution will be-
A. 2
B. 7
C. 5
D. 1

Answer 1

Hint: In this solution, we will simply find out the total volume of the liquid and the number of ${H^ + }$ ions present in the liquid because as we all know that the hydrogen ion level in moles per litre (molarity) must be known in order to calculate the pH of an aqueous solution.

Formula used: $pH = - \log \left[ {{H^ + }} \right]$

Complete step-by-step answer:
As already given in the question-
$\Rightarrow HCl = 0.01N(1mL) \\\ \\\ \Rightarrow N{a_2}S{O_4} = 0.1N(999mL) \\\$
pH of any solution always depends on the concentration of ${H^ + }$ ion in the solution. So, the concentration of ${H^ + }$ ions in the above solution is-
$\Rightarrow {H^ + } = 0.01mol$ in 1mL
Total volume of the solution if calculated from the given information is-
$\Rightarrow 1 + 999 = 1000mL$
Thus, the number of ${H^ + }$ ions in the above solution in terms of N is-
$\Rightarrow \dfrac{{0.01}}{{1000}} = {10^{ - 5}}N$
The formula used is $pH = - \log \left[ {{H^ + }} \right]$. Substituting the values in the given formula, we will have-
$\Rightarrow pH = - \log \left[ {{H^ + }} \right] \\\ \\\ \Rightarrow pH = - \log \left[ {{{10}^{ - 5}}} \right] \\\ \\\ \Rightarrow pH = 5 \\\$
Hence, option C is the correct option.

Note: pH, functional acidity analysis or basic acidity or other liquid solutions. The word, which is commonly used in chemical, biological and agronomic fields, converts hydrogen ion concentration values — typically between 1 and ${10^{ - 14}}$ g-equivalents per litre — into numbers from 0 to 14. The concentration of the hydrogen ion is ${10^{ - 7}}$ gram-equivalent per litre of liquid water, stable (neither alkaline nor acidic), which correlates to a pH of 7. A pH less than 7 solution is called acidic; pH greater than 7 solution is known to be basic, or alkaline.