A. Assuming the density of a 5% acetic acid is 1.0 g/mL,determine the volume of

Question

A. Assuming the density of a 5% acetic acid is 1.0 g/mL,determine the volume of the acetic acid solution necessary tonutralize 25.0 mL of 0.10 M NaOH. B. A 31.43 mL colume of 0.108 M NaOH is required to reach thephenolphthlein endpoint in the titration of a 4.441 g sample ofvinegar. Caculate the percent of acetic acid in vinegar. C. Lemon juice has a pH of about 2.5. Assuming that theacidity of a lemon juice is due soley to citric acid, that citricacid is a monoprotic acid, and that the density of lemon juice is1.0 g/mL, then the citric acid concentration caculates to 0.5% bymass. Estimate the volume of 0.0100 M NaOH required to nutralize a3.71 g sample of lemon juice. The molar mass of citric acid is190.12 g/mol/ A. Assuming the density of a 5% acetic acid is 1.0 g/mL,determine the volume of the acetic acid solution necessary tonutralize 25.0 mL of 0.10 M NaOH. B. A 31.43 mL colume of 0.108 M NaOH is required to reach thephenolphthlein endpoint in the titration of a 4.441 g sample ofvinegar. Caculate the percent of acetic acid in vinegar. C. Lemon juice has a pH of about 2.5. Assuming that theacidity of a lemon juice is due soley to citric acid, that citricacid is a monoprotic acid, and that the density of lemon juice is1.0 g/mL, then the citric acid concentration caculates to 0.5% bymass. Estimate the volume of 0.0100 M NaOH required to nutralize a3.71 g sample of lemon juice. The molar mass of citric acid is190.12 g/mol/

Explanation / Answer

molarity of acetic acid solution: 5% = 50g/litre
Molar mass acetic acid = 60.05 g mol1
Molarity = 50/60.05 = 0.833M

Balanced equation = CH3COOH + NaOH CH3COONa + H2O
Molar ratio = 1:1
Use equation:
M1V1 = M2V2
0.833 *V1 = 25*0.1
V1 = 2.5/0.833
V1 = 3.00ml

Volume of acetic acid required = 3.0ml

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B. A31.43 mL colume of 0.108 M NaOH is required to reach thephenolphthlein endpoint in the titration of a 4.441 g sample ofvinegar. Caculate the percent of acetic acid invinegar.

moles ofsolute = M * V
0.108 M NaOH * 0.03143 = 0.00339 moles of solute NaOH
moles of acetic acid = moles of NaOH
weight of acid = 60.05 g/mol * 0.00339 = 0.204 grams
0.204 g / 4.441 g * 100 = 4.60 %

---------------------------------------C.

Lemon juice has a pH of about 2.5. Assuming that theacidity of a lemon juice is due soley to citric acid, that citricacid is a monoprotic acid, and that the density of lemon juice is1.0 g/mL, then the citric acid concentration caculates to 0.5% bymass. Estimate the volume of 0.0100 M NaOH required to nutralize a3.71 g sample of lemon juice.

The molar mass of citric acid is 190.12g/mol/

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