A 0.3240g sample of impure #Na_2CO_3# was dissolved in 50.00mL of 0.1280M #HCl#. The excess acid then requires 30.10mL of 0.1220M #NaOH# for complete neutralization. How do you calculate the % #Na_2CO_3# (MM = 105.99) in the sample?

2 Answers
Jul 2, 2016

Answer:

You can do it like this:

Explanation:

Sodium carbonate reacts with hydrochloric acid:

#sf(Na_2CO_(3(s))+2HCl_((aq))rarr2NaCl_((aq))+CO_(2(g))+H_2O_((l)))#

  • We can find the number of moles of #"HCl"# before the reaction.

  • We then use the titration result to find the number of moles remaining after the reaction.

  • By subtracting the two we can get the number of moles of #"HCl"# which have reacted.

  • From the equation we can find the number of moles of #"Na"_2"CO"_3#.

  • From this we get the mass of #"Na"_2"CO"_3#.

  • We can then work out the percentage purity.

Concentration = moles of solute / volume of solution.

#c=n/v#

#:.n=cxxv#

#:.# Initial moles of #sf(HCl"=0.1280xx50.00/1000=6.400xx10^(-3))#

The acid remaining is titrated with #"NaOH"#:

#sf(HCl_((aq))+NaOH_((aq))rarrNaCl_((aq))+H_2O_((l)))#

#sf(nOH^(-)=0.1220xx30.10/1000=3.6722xx10^(-3))#

Since they react in a 1:1 ratio the no. of moles of #"HCl"# must be the same:

#sf(nHCl=3.6722xx10^(-3))#

#:.# The no. moles used up:

#sf(=(6.400-3.6722)xx10^(-3)=2.7228xx10^(-3))#

From the original equation you can see that the no. moles of #"Na"_2"CO"_3# must be half of this.

#:.sf(nNa_2CO_3=(2.7228xx10^(-3))/2=1.3614xx10^(-3))#

#sf(M_r[Na_2CO_3]=105.99)#

#:.# #sf(mass" "Na_2CO_3=1.3614xx10^(-3)xx105.99=0.14429" "g)#

#:.# #sf("percentage purity"" "=(0.14429)/(0.3240)xx100=44.53%)#

This technique is known as a "back titration".

Jul 4, 2016

Answer:

#~~44.61%#

Explanation:

The equivalent mass of #Na_2CO_3#
#=("molar mass of "Na_2CO_3)/"total valency of metal"#

#=105.99/2=52.995g/"equivalent"#

Let the amount of #Na_2CO_3# in 0.3240g sample be #(x g)=x/52.995"g.equivalent"#

In HCl and NaOH the molar mass and eqivalent mass are same.So for their solutions molarity is same as normality.

So 50 mL 0.1280M or 0.1280N HCl solution will contain #(50xx0.1280)/1000 "g.equivalent"# of HCl

And 30.10mL 0.1220M or N NaOH solution will contain #(30.1xx0.1220)/1000"g.eqivalent"# NaOH
Now by the law of equivalent proportion nutralisation will occur if total no.of g.eqivalent of base is same as total no.of g.equivalent of acid.

Hence

#x/52.995+(30.1xx0.122)/1000=(50xx0.1280)/1000#

#x=(50xx0.128-30.1xx0.122)/1000xx52.995#
#~~0.14456 g#

#"% of purity "=(0.14456/0.324)xx100~~44.61#