Question

2g of a mixture sodium hydroxide and sodium chloride (as impurity) were dissolved in 500cm³ of water.if 25cm³ of this solution were neutralized by 21 cm³ 0.1dm-³ hydrochloric acid calculate the percentage of the sodium chloride impurity

Answer 1

Equation of reaction is NaOH_(aq)+HCl_(aq)=NaCl_(aq)+H₂O_(l)

Moles of HCl that reacted

moles=molarity*volume/1000

=0.1*21/1000

=0.0021 moles of HCl

Since the mole ratio between NaOH : HCl ,

moles of NaOH that reacted is 0.0021 moles.

If 25mL contains 0.0021 moles of NaOH

500mL solution contain?

500ml*0.0021moles/25mL

=0.042moles

Mass of NaOH dissolved in 500mL

=40gmol^-1*0.042moles

=1.68g

mass of NaCl in 500mL solution

=(2-1.68)g

=0.32g

% of NaCl impurity= (0.32/2)*100

=16% NaCl