Question

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?

Answer 1

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".

Answer 2

`~~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`