Question

A $10.03$g of vinegar was diluted to 100 mL and 25 mL sample was titrated with the $0.0176$M ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$solution. $34.30$mL was required for equivalence. What is the percentage of acetic acid in the vinegar?

Answer 1

Hint:This question is based on molarity, which is defined as the number of moles of a substance present per litre of the solution and one mole is defined as $6.022 \times {10^{23}}$particles, where particles can be atoms, molecules, ions and subatomic particles.
Formula Used: \[{\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}\]

Complete step by step answer:
According to the question,
The amount of vinegar present in 100 mL = $10.03$g
The volume of vinegar taken for titration = 25 mL
Volume of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ solution required for titration of acetic acid = $34.30$mL
The molarity of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ solution = $0.0176$M
Now, the normality of the solution = \[{{n - factor \times molarity}}\],
where n-factor is the number of equivalents in gram. Here, for ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$, n = 2. Hence,
Normality of the solution = \[{{2 \times }}0.0176\]= $0.0352$N.
From the formula of the normality, we know that,
${{\text{V}}_{\text{1}}}{\text{ }}{{\text{S}}_{\text{1}}}{\text{ = }}{{\text{V}}_{\text{2}}}{\text{ }}{{\text{S}}_{\text{2}}}$, here, ${{\text{V}}_{\text{1}}}$= 25 mL, ${{\text{S}}_{\text{1}}}$is not known, ${{\text{V}}_2}$= $34.30$mL, and ${{\text{S}}_2}$= $0.0352$N.
Therefore,
${\text{ }}{{\text{S}}_{\text{1}}}{\text{ = }}\dfrac{{{{\text{V}}_{\text{2}}}{\text{ }}{{\text{S}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$
\[ \Rightarrow {\text{ }}{{\text{S}}_{\text{1}}}{{ = }}\dfrac{{34.30 \times {\text{ }}0.0352}}{{{\text{25}}}} = 0.0482\]
Hence, the normality of acetic acid is $0.0482$N. Now, to convert the normality of acetic acid to its molarity, we divide the normality by the no. of equivalents of acetic acid.
Hence, molarity is,
\[\dfrac{{0.0482}}{1}\]=\[0.0482\]M.
Hence,\[0.0482\]moles of acetic acid are present per litre of the solution or in 1000 mL of the solution. Hence, the amount of acetic acid present in 100 mL of the solution = \[0.00482\] moles.
The molecular mass of acetic acid = $\left[ {12 + \left( {3 \times 1} \right) + 12 + \left( {16 \times 2} \right) + 1} \right] = 60$
Hence 1 mole of acetic acid = $60$g
Therefore, \[0.00482\]moles of acetic acid = \[0.00482 \times 60 = 0.2892\]g.
The percentage of acetic acid in vinegar = $\dfrac{{{\text{mass of acetic acid}}}}{{{\text{mass of vinegar}}}}{{ \times 100}}$
$ \Rightarrow \dfrac{{{\text{0}}{\text{.2892}}}}{{{\text{10}}{\text{.03}}}}{{ \times 100}}$= $2.88$%.

So, the percentage of acetic acid in vinegar is $2.88$%.

Note:
The number of gram equivalents of any substance present in one litre or 1000 mL of the solution is defined as the normality of the solution while the percentage composition of a solution is defined as the number of grams of the solute present in 100 mL of the solution.