Question

How many grams of sodium bicarbonate are required to neutralise 20 ml of 0.902m vinegar?

Answer 1

vinegar is the common name of acetic acid ( CH3COOH).

now, sodium bicarbonate (NaHCO3) is neutralised by 20ml of 0.902M acetic acid.

no of mole of acetic acid = concentration of acetic acid × volume of solution in L

= 0.902M × 20/1000

= (0.902/50) = 0.01804 mol

so, equivalents of sodium bicarbonate = equivalents of acetic acid

or, weight of NaHCO3/equivalent weight of NaHCO3 = no of mole of acetic acid/n-factor of acetic acid

or, weight of NaHCO3/(molar weight of NaHCO3/n - factor of NaHCO3) = 0.01804/1

or, weight of NaHCO3/(84/1) = 0.01804

or, weight of NaHCO3 = 84 × 0.01804 = 1.51536g