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.
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
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