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

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

Expert Solution

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

= 2.2939 x 10^-3 / 3

= 7.65 x 10^-4

molar mass of citric acid = 192.12 g/mol

mass 0f citric acid = moles x molar mass

= 7.65 x 10^-4 x 192.12

= 0.147 g

% citric acid = (0.147 / 2.00 ) x 100

= 7.34%

queen_honey_blossom answered 2 years ago

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