1 answer

In determination of percent purity of sodium carbonate a 25cm³ aliquot of a 5.4g/l sample required 17.57cm³ of 0.07495m soln

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?

Answers

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#

.

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