A 2.00 g sample of a powdered fruit drink containing citric acid, required 20.3 Ml of...

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A 2.00 g sample of a powdered fruit drink containing citricacid, required 20.3 Ml of 0.113 M NaOH for complete neutralization.calculate the percent citric acid in the sample, Please show yourwork so i can understand.

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3.8 Ratings (668 Votes)

C3H5O(COOH)3 (aq) + 3 NaOH(aq) --------------------> Na3C3H5O(COO)3(aq) + 3 H2O (l)

moles of NaOH = 20.3 x 0.113 / 1000 = 2.2939 x 10^-3

here

1 mol citric acid react -------------- 3 moles of NaOH

so moles of citric acid = moles of NaOH / 3

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.2939 x 10^-3 / 3

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.65 x 10^-4

molar mass of citric acid = 192.12 g/mol

mass 0f citric acid = moles x molar mass

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.65 x 10^-4 x 192.12

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.147 g

% citric acid = (0.147 / 2.00 ) x 100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.34%

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A 2.00 g sample of a powdered fruit drink containing citric acid, required 20.3 Ml of...

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50.1K

Verified Solution

Question

Chemistry

A 2.00 g sample of a powdered fruit drink containing citricacid, required 20.3 Ml of 0.113 M NaOH for complete neutralization.calculate the percent citric acid in the sample, Please show yourwork so i can understand.

Answer & Explanation Solved by verified expert

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