Question

Two solutions, A and B, each of $100{\text{L}}$ was made by dissolving $4{\text{g}}$ of ${\text{NaOH}}$ and $9.8{\text{g}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in water, respectively. The pH of the resultant solution obtained from mixing $40{\text{L}}$ of solution A and $10{\text{L}}$ of solution B is ______.

Answer 1

Hint: To solve this question, you must use the concept of equivalents. Equal numbers of equivalents of an acid neutralize equal numbers of equivalents of a base. We shall calculate the molarity of the acid and base present and thus, the molarity of acid/base present after neutralization. The molarity of hydrogen ion or hydroxide ion present in the solution after neutralization shall be used to calculate the pH of the solution.
Formula used:
 ${{meq}} = {{M \times V \times }}x$
Where ${\text{meq}}$ is the number of milliequivalents.
M is the molarity of the acid / base solution
V is the volume of the solution
x is the n-factor of the acid or base

Complete step by step answer:
We know that, the acidity of${\text{NaOH}}$ is equal to 1 while the basicity of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$is equal to 2, which are also the respective n factors of the two compounds.
The number of moles of a substance are equal to the ratio between its given mass and its molar mass.
The number of moles of ${\text{NaOH}}$ are, ${\text{n}} = \dfrac{4}{{40}} = 0.1$
The molarity is given as the number of moles per litre of the solution.
Molarity of ${\text{NaOH}}$ solution $ = \dfrac{{0.1}}{{100}} = {10^{ - 3}}{\text{M}}$
The number of moles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are, $\dfrac{{9.8}}{{98}} = 0.1$
The molarity of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$solution$ = \dfrac{{0.1}}{{100}} = {10^{ - 3}}{\text{M}}$
$40{{L}}$ of solution A will contain $4 \times {10^{ - 2}}$ moles of ${\text{NaOH}}$
$10{{L}}$ of solution B will contain ${10^{ - 2}}$ moles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
The equivalents of ${\text{NaOH}}$ and ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are $4 \times {10^{ - 2}}$ and $2 \times {10^{ - 2}}$ respectively.
After neutralization, $2 \times {10^{ - 2}}$ equivalents of ${\text{NaOH}}$ remain.
$\left[ {{\text{O}}{{\text{H}}^ - }} \right] = \dfrac{{2 \times {{10}^{ - 2}}}}{{50}} = 4 \times {10^{ - 4}}{\text{N}}$
As a result we can write, $\left[ {{{\text{H}}^ + }} \right] = \dfrac{{{{10}^{ - 14}}}}{{4 \times {{10}^{ - 4}}}} = 2.5 \times {10^{ - 11}}{\text{N}}$
The pH of the solution will be equal to
\[{\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right] = - \log \left( {2.5 \times {{10}^{ - 11}}} \right)\]
$\therefore {\text{pH}} = 10.6$

Note:
The n factor is also known as the valence factor and its value varies depending upon the compound being considered. For an acid, the n factor is the basicity of the acid. Similarly, the n factor of a base is its acidity. For electrolytes, it is the valency of the ions in the solution.