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?

Question

Ellie Andrews · Accepted Answer

#~~44.61%#

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#

Peyton Adams · Answer

You can do it like this:

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".

Caleb Adams · Answer

undefined First, calculate the moles of HCl used. Then, use the stoichiometry of the reaction between HCl and Na2CO3 to find the moles of Na2CO3. Finally, divide the moles of Na2CO3 by the total mass of the sample and multiply by 100 to get the percentage of Na2CO3 in the sample.