Question

A major organic acid in this juice is malic acid, a dicarboxylic acid with molecular weight of 134. You are asked to determine the total acidity of this juice and a 10-mL portion of this juice took 8.5 mL of 0.105 N NaOH to neutralize to the phenolphthalein endpoint. What is the acidity in mg malic acid/mL juice unit and in % (weight) unit?

Answer 1

Concentration of NaOH = 0.105 N = 0.105 M (because molar mass of NaOH equals equivalent weight of NaOH)

volume of NaOH = 8.5 mL = 0.0085 L

moles of NaOH = (concentration of NaOH) * (volume of NaOH in Liters)

moles of NaOH = (0.105 M) * (0.0085 L)

moles of NaOH = 8.925 x 10^-4 mol

Since malic acid is diprotic acid, therefore, each molecule of malic acid requires two molecules of NaOH for complete neutralization.

moles of malic acid = (moles of NaOH) / 2

moles of malic acid = (8.925 x 10^-4 mol) / 2

moles of malic acid = 4.4625 x 10^-4 mol

mass of malic acid = (moles of malic acid) * (molar mass of malic acid)

mass of malic acid = (4.4625 x 10^-4 mol) * (134 g/mol)

mass of malic acid = 0.0598 g

mass of malic acid = 59.8 mg

acidity = (mass of malic acid) / (volume of juice)

acidity = (59.8 mg) / (10 mL)

acidity = 5.98 mg/mL juice

Assuming density of juice = 1.00 g/mL

mass of juice = (volume of juice) * (density of juice)

mass of juice = (10 mL) * (1.00 g/mL)

mass of juice = 10 g

% weight = (mass of malic acid / mass of juice) * 100

% weight = (0.0598 g / 10 g) * 100

% weight = 0.598 %