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Kohlrausch Law

10.03 g of vi...

Question

10.03 g of vinegar was diluted to 100 mL and 25 mL sample was titrated with the 0.0176 M $B a (O H)_{2}$ solution. 34.30 mL was required for equivalence. What is the percent of acetic acid in the vinegar?

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Solution

Given, Mass of Vinegar= $10.03 g$
Molarity of $B a (O H)_{2} = 0.0176 M$
Volume of vinegar solution used= $25 m l$
Volume of $B a (O H)_{2}$ used= $34.30 m l$
We know, at equivalence, no. of gram equivalents of vinegar solution=No. of gram equivalents of $B a (O H)_{2} ⟶ (1)$
Now,
Molarity of $B a (O H)_{2} = 0.0176 M$
$\Rightarrow$ Normality of $B a (O H)_{2} = 0.0176 \times n f a c t o r$
$= 0.0176 \times 2$
$= 0.0352 N$
So, from $(1)$ $N_{1} V_{1} = N_{2} V_{2}$
$V i n e g a r s o l .$ $B a (O H)_{2}$
$\Rightarrow N_{1} = \frac{N_{2} V_{2}}{V_{1}} = \frac{0.0352 \times 34.30}{25}$
$= 0.0482 N$
$∴$ The normality of acetic acid in the vinegar solution is $0.0482 N$
Molarity of acetic acid= $\frac{0.0482}{n f a c t o r}$ $[∵ n = 1]$
$= \frac{0.0482}{1} M$
This means that
$0.0482$ moles of acetic acid is present in $1000 m l$ solution
$\Rightarrow$ No. of moles of acetic acid in $100 m l$ solution is= $\frac{0.0482 \times 100}{1000}$
$= 0.00482$ moles
Now, for % calculation , first let us calculate the mass
We know $1$ mole acetic acid= $60 g$
$\Rightarrow 0.00482$ mole acetic acid= $0.00482 \times 60$
$= 0.2892 g$
Now, for % calculation
$∴$ Mass % of acetic acid in vinegar solution= $\frac{M a s s o f A c i d}{M a s s o f V i n e g a r} \times 100$
$= \frac{0.2892 g}{10.03 g} \times 100$
$= 2.88$ %

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10.03 g of vinegar was diluted to 100 mL and 25 mL sample was titrated with the 0.0176 M Ba(OH)2 solution. 34.30 mL was required for equivalence. What is the percent of acetic acid in the vinegar?

10.03 g of vinegar was diluted to 100 mL and 25 mL sample was titrated with the 0.0176 M $B a (O H)_{2}$ solution. 34.30 mL was required for equivalence. What is the percent of acetic acid in the vinegar?