Experiment 10 Vinegar Analysis Vinegar is o 4-5% by mass) souon in ocesc ocid th

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Experiment 10 Vinegar Analysis Vinegar is o 4-5% by mass) souon in ocesc ocid the percent by mass of acetic acid in vinegar . To detemine the Jliswing techaiques are used in the Experimental Pocedure OBJECTIVE The following techniques are used in TECHNIQUES 164 Household vinegar is a 4-5% nitinum foderal o make the product is a 4-5% (by mass) acetic acid, CHCOOH, solution (4% is the standard). Generally, caramel flavoring and coloring are also added 'NTRODUCTION make the product acsthetically more appealing. make the pene anaysis using the titration technique is the method used for determin- A volumetric he percent by mass of acetic acid in vincgar. A measured mass of vinegar is titrated n endpoint with a measured volume of a standardized sodium o the olution. Since the volume and molar concentration of the standardized NaOH known, th acetic acid w he moles of NaOH used for the analysis are also known. mol NaOH-L NaOH solution × mol NaOH L NaOH solution From the balanced equation: CH,COOH(ag) + NaOHag)NaCH,CO,(a)+HO)(10.2) (10.3) The mass of CH,COOH in the vinegar is calculated from the measured moles of mol CH,COOH mol NaOH CH,COOH neutralized in the reaction and its molar mass,60.05 g/mol: 60.05 g CH COOH CH.co0H × eoes g CHCOOH H 104) mol CH,COOH(10.4) mass(g) of CHCOOH-mol CH3COOH × Finally, the percent by mass of CH,COOH in vinegar is calculated: % by mass of CH,COOH = masig)of CHCOOH x 100 (10.5) mass (g) of vinegar Procedure Overview: Samples of one or two vinegars are analyzed for the amount rclc acid in the sample. A titration setup is used for the analysis, using a standard- ned NaOH solution as the titrant and EXPERIMENT PROCEDURE NO solution as the titrant and phenolphthalein as the indicator Experiment

Explanation / Answer

Given that ; density of 5% solution vinegar = 1.0 g/ ml

Volume of NaOH = 25 ml

Molarity = 0.10 M

First we write the recation between CH3COOH and NaOH:

CH3COOH + NaOH = CH3COONa + H2O

Moles of NaOH= 0.10* 0.025

= 2.5*10^-3 Moles NaOH

According to reaction 1 mole of acetic acid reacts with one mole of base

Moles of acid = 2.5*10^-3 Moles.

Assume that the total volume of 5% vinegar solution is 100 ml.

Mass of solution = volume * density

= 100 ml* 1.0 g/ ml

= 100 g

molar mass of acetic acid= 60.05 g/mol

Amount of CH3COOH = 5 g

Moles of CH3COOH = 5.0 g/ 60.05 g/mol

= 0.083 moles CH3COOH

Here 0.083 moles CH3COOH present in 100 ml

2.5*10^-3 Moles CH3COOH will present

= 100 ml /0.083 mole * 2.5*10^-3

= 3.012 ml

Thus we need 3.0 ml of 5% acetic acid to neutralize 25 ml of 0.100 M NaOH solution.

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