Question

In determination of percent purity of sodium carbonate a 25cm³ aliquot of a 5.4g/l sample required 17.57cm³ of 0.07495m soln of dilute hcl acid to reach methyl orange end point. calculate the percentage purity of the sodium carbonate sample?

Answer 1

Strength of given `HCl` solution is `0.07495m`

Volume of given `HCl` solution required for neutralization of `25mL` aliquot of `Na_2CO_3` solution is `17.57mL=0.01757L`

So this acid solution contains `0.07495xx0.01757mol` HCl

Now the equation of neutralization reaction is

`2HCl+Na_2CO_3->2NaCl+H_2O+CO_2`.
This equation suggests that 1mol HCl requires 0.5mole `Na_2CO_3` for neutralization.

Hence the amount of pure `N_2CO_3` present in `25mL` aliquot is `0.07495xx0.01757xx0.5mol`

Hence g/L strength of pure `Na_2CO_3` is

`0.07495xx0.01757xx0.5xx106xx1000/25` g/L.

Hence percentage of purity will be

`=0.07495xx0.01757xx0.5xx106xx1000/25xx100/5.4` ℅

`~~51.7`℅