Percent purity

Question

A 0.347 g sample of impure NaHSO₃ is dissolved in water and titrated with 0.1481 M NaOH. If 20.20 mL of NaOH is required, what is the percentage purity of the NaHSO₃ sample?

Answer 1

Balanced equation for the titration is:

NaOH + NaHSO3 => H2O + Na2SO3

moles NaOH used = 0.1481 mol/L x0.02020 L = 0.002992 moles

 

moles of pure NaHSO3 = 0.002992 mokes based on 1:1 mole ratio in balanced equation

mass NaHSO3 = 0.002992 mol x 104.1 g/mol = 0.311g

 

% purity 0.311g/0.347g (x100%) = 89.7%