Question

A solution of tartaric acid is prepared of concentration 1.2174 g L -1 and it is found that 50.0 ml of the solution requires 21.48 mL of 3.78 × 10-2 mol L -1 sodium hydroxide for complete neutralization. The formula of tartaric acid is C4H6O6 and its relative molecular mass is 150.2. Calculate how many of the hydrogens in a molecule of tartaric acid are ionised in aqueous solution.

Answer 1

Given : mass concentration of tartaric acid = 1.2174 g/L

volume of tartaric acid solution = 50.0 mL = 0.0500 L

mass tartaric acid = (mass concentration of tartaric acid) * (volume of tartaric acid solution)

mass tartaric acid = (1.2174 g/L) * (0.0500 L)

mass tartaric acid = 0.06087 g

moles tartaric acid = (mass tartaric acid) / (molar mass tartaric acid)

moles tartaric acid = (0.06087 g) / (150.2 g/mol)

moles tartaric acid = 4.05 x 10^-4 mol

concentration NaOH = 3.78 x 10^-2 mol/L

volume NaOH required = 21.48 mL = 0.02148 L

moles NaOH = (concentration NaOH) * (volume NaOH required)

moles NaOH = (3.78 x 10^-2 mol/L) * (0.02148 L)

moles NaOH = 8.12 x 10^-4 mol

number of hydrogen = (moles NaOH) / (moles tartaric acid)

number of hydrogen = (8.12 x 10^-4 mol) / (4.05 x 10^-4 mol)

number of hydrogen = 2