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A beverage mixture contains citric acid, C₆H₈O₇ (molar mass 192 g/mol), plus sugar and flavorings. The mass of citric acid in a sample was determined by titration with sodium hydroxide solution. If a 1.500 g sample of the beverage mix required 46.65 mL of 0.1068 N NaOH for complete neutralization, how many grams of citric acid were initially present in the sample?

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Answer:

0.319 gram

Step-by-step explanation:

Convert Normality to Molarity.

for NaOH: 0.1068 N = 0.1068 M

Citric acid contains 3 groups of COOH so that 1 mole of citric acid reacts with 3 moles of NaOH.

#moles of NaOH = Concentration x Volume = 0.1068 x 46.65 x 10^-3

= 4.98 x 10^-3 mol

#moles of Citric acid = #moles of NaOH/3 = 1.66 x 10^-3 mol

mass of citric acid = mole x molar mass = 1.66 x 10^-3 x 192 = 0.319 gram

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