Question

Find the pH of solution ${\text{300 ml of 1}}{{{0}}^{{{ - 4}}}}{{ M Ba(OH}}{{{)}}_{{2}}}$ and ${{500 ml of 2 \times 1}}{{{0}}^{{{ - 5}}}}{{ M }}{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}$?

Answer 1

First we will calculate the normality of the solutions of both the acid and bases and calculate the amount of protons and hydroxide ions released in the solution then calculate the pH by the formula of the pH.

Complete step by step answer:
As we know pH is the amount of protons in the solution as to how much acidic is the solution. But when there is a mixing of the acid and base there is neutralization of ions occurring. The released protons and hydroxide ions neutralize each other and the ones present in more quantity are left in the solution and decide the acidity or basicity of the solution.
The formula for the final normality is
${{{N}}_{{{mix}}}}{{ = }}\dfrac{{{{(}}{{{N}}_{{{Ba(OH}}{{{)}}_{{2}}}}}{{ \times }}{{{V}}_{{{Ba(OH}}{{{)}}_{{2}}}}}{{) - (}}{{{N}}_{{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}}}{{ \times }}{{{V}}_{{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}}}{{)}}}}{{{{(}}{{{V}}_{{{Ba(OH}}{{{)}}_{{2}}}}}{{ + }}{{{V}}_{{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}}}{{)}}}}$
First we will find the normality of both the solutions. Normality, ${{N}}$ can be calculated by multiplying molarity, ${{M}}$ with z-factor, ${{z}}$.
${{N = M \times z}}$
Since both the acid and base are diprotic and dibasic so the z-factor is $2$ for both
${{{N}}_{{{Ba(OH}}{{{)}}_{{2}}}}}{{ = 1}}{{{0}}^{{{ - 4}}}}{{ \times 2}}$
${{{N}}_{{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}}}{{ = 2 \times 1}}{{{0}}^{{{ - 5}}}}{{ \times 2}}$
Volume should be in liters so convert the ml into liter and substitute in final normality formula
${{{N}}_{{{mix}}}}{{ = }}\dfrac{{{{(2 \times 1}}{{{0}}^{{{ - 4}}}}{{ M \times 0}}{{.3 L) - (4 \times 1}}{{{0}}^{{{ - 5}}}}{{ M \times 0}}{{.5 L)}}}}{{{{(0}}{{.3 + 0}}{{.5) L}}}}$
${{{N}}_{{{mix}}}}{{ = 6}}{{.25 \times 1}}{{{0}}^{{{ - 5}}}}$
Now this calculated normality is the value for amount of ${{O}}{{{H}}^{{ - }}}$ left in solution because the normality of base was higher than the normality of the acid. So we will calculate ${{pOH}}$
${{pOH = - log [}}{{{N}}_{{{mix}}}}{{]}}$
$pOH = - \log [6.25 \times {10^{ - 5}}]$
${{pOH = 4}}{{.2041}}$
Since we know that ${{pH + pOH = 14}}$
So ${{pH = 14 - 4}}{{.2041 = 9}}{{.7959}}$
And hence the answer is $9.8$.

Additional Information:
We should use and calculate normality instead of molarity as it is more precise and less calculation will be there. Normality will be the same and just depends on z-factor and there will be less mistakes in calculating out the moles in solution of mono-protic , di-protic , tri-protic acid without any chemical equation.

Note:
It should be paid attention to the number and type of ions left in the solution. If the base had more ions then the solution will have hydroxide ions more and the solution will be basic. If the acid has more ions than protons will be more in the solution and the solution will be acidic.