Question

Calculate the pH of the solution obtained by mixing 10 mL of $0.1{\text{M HCl}}$ and 40 mL of $0.2{\text{M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$

Answer 1

To solve this question, first we shall find the number of moles of $[{{\text{H}}^ + }]$ ions present in ${\text{HCl}}$ and in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ separately which we shall do using the relation between molarity, moles of solute and volume of the solution. We will then add them to get the total number of moles of $[{{\text{H}}^ + }]$ ions present in the solution from which we will calculate the concentration of $[{{\text{H}}^ + }]$ ions in the mixture. This molarity can then be used to find the pH of the given solution.

Formula Used: ${\text{Molarity = }}\dfrac{{{\text{number of moles}} \times {\text{1000}}}}{{{\text{volume of solution(in mL)}}}}$ and ${\text{pH}} = - \log [{{\text{H}}^ + }]$
Where $[{{\text{H}}^ + }]$ is the concentration of hydrogen ions present in the solution.

Complete step-by-step answer:
We will calculate the number of moles of $[{{\text{H}}^ + }]$ ions present in $0.1{\text{M HCl}}$ and $0.2{\text{M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ respectively by using the information given to us.
${\text{HCl}}$in the solution will form:
$HCl \to {H^ + } + C{l^ - }$
As only 1 mole of $[{{\text{H}}^ + }]$ ion is formed, molarity of $[{{\text{H}}^ + }]$ ion is equal to the molarity of ${\text{HCl}}$, i.e. $0.1{\text{M}}$.
So, for 10 mL of $0.1{\text{M HCl}}$:
Let the number of moles of $[{{\text{H}}^ + }]$ ion in ${\text{HCl}}$ be ${\text{n}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$.
We know that, ${\text{Molarity = }}\dfrac{{{\text{number of moles}} \times {\text{1000}}}}{{{\text{volume of solution(in mL)}}}}$.
Substituting the value of molarity of ${\text{HCl}}$ in the formula and rearranging it, we get:
$\Rightarrow {\text{0}}{\text{.1 = }}\dfrac{{{\text{n}}{{\text{'}}_{[{{\text{H}}^ + }]}} \times {\text{1000}}}}{{{\text{10}}}}$
$\Rightarrow {\text{n}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} = \dfrac{{0.1 \times 10}}{{1000}}$
Solving this for ${\text{n}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$ we get,
${\text{n}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} = {10^{ - 3}}$ moles of $[{{\text{H}}^ + }]$ in ${\text{HCl}}$.
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in the solution will form:
${H_2}S{O_4} \to 2{H^ + } + SO_4^{2 - }$
As 2 moles of $[{{\text{H}}^ + }]$ ion is formed, the molarity with respect to $[{{\text{H}}^ + }]$ ions will be twice the molarity of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, i.e. $0.2 \times 2 = 0.4{\text{M}}$.
So, for $0.2{\text{M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$:
Let the moles of $[{{\text{H}}^ + }]$ ion in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ be ${\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$.
We know that, ${\text{Molarity = }}\dfrac{{{\text{number of moles}} \times {\text{1000}}}}{{{\text{volume of solution(in mL)}}}}$.
Substituting the value of molarity of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in the formula and rearranging it, we get:
$\Rightarrow {\text{0}}{\text{.4 = }}\dfrac{{{\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} \times {\text{1000}}}}{{{\text{40}}}}$
$\Rightarrow {\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} = \dfrac{{0.4 \times 40}}{{1000}}$
Solving this for ${\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$we get,
${\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} = 16 \times {10^{ - 3}}$moles of $[{{\text{H}}^ + }]$ in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
So, the total moles of $[{{\text{H}}^ + }]$ ions is:
${{\text{n}}_{{\text{total}}}} = {\text{n}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}} + {\text{n'}}{{\text{'}}_{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$
$\Rightarrow {{\text{n}}_{{\text{total}}}} = 1 \times {10^{ - 3}} + 16 \times {10^{ - 3}}$
Solving this:
$\Rightarrow {{\text{n}}_{{\text{total}}}} = 17 \times {10^{ - 3}}$
Now, we shall find the molarity of $[{{\text{H}}^ + }]$ ions in the mixture of ${\text{HCl}}$ and ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
${\text{Molarity = }}\dfrac{{{{\text{n}}_{{\text{total}}}} \times {\text{1000}}}}{{{\text{volume of solution (in mL)}}}}$
Substituting the value of ${{\text{n}}_{{\text{total}}}}$ and the volume of solution, we get:
$\Rightarrow {\text{Molarity}} = \dfrac{{17 \times {{10}^{ - 3}} \times 1000}}{{10 + 40}}$
Solving this equation, we get:
$\Rightarrow {\text{Molarity}} = \dfrac{{17}}{{50}}$
$\Rightarrow {\text{Molarity = }}0.34$
The pH of a solution is given by: ${\text{pH}} = - \log [{{\text{H}}^ + }]$
Substituting the value we get:
$\Rightarrow {\text{pH}} = - \log [0.34]$
Solving this to get the pH of the solution we get,
$\Rightarrow {\text{pH}} = - (0.47)$
${\text{pH}} = 0.47$
So, the pH of the given solution of 10 mL of $0.1{\text{M HCl}}$ and 40 mL of $0.2{\text{M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $0.47$ .

Note: pH of a solution only depends upon the concentration of $[{{\text{H}}^ + }]$ ions present in the solution, it is unaffected by the anion present. For strong acids, dissociation constant is ideally taken as 1. So, the molar concentration of $[{{\text{H}}^ + }]$ ions in the solution will be dependent only on the stoichiometric coefficient of the dissociation reaction and the molar concentration of the acid.