Answer:
The concentration of acetic acid in the vinegar is 7,324 (%V/V)
Explanation:
The titration equation of acetic acid with NaOH is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
The moles required were:
1,024M×0,02500L = <em>0,02560 moles NaOH. </em>These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:
0,02560 moles CH₃COOH×= <em>1,537g of CH₃COOH</em>
As density is 1,01g/mL:
1,537g CH₃COOH×= <em>1,522mL of CH₃COOH</em>
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As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
×100= <em>7,324 (%V/V)</em>
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I hope it helps!
