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
Answer:
You can do it like this:
Explanation:
Sodium carbonate reacts with hydrochloric acid:

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.
The acid remaining is titrated with
Since they react in a 1:1 ratio the no. of moles of
From the original equation you can see that the no. moles of
This technique is known as a "back titration".
Answer:
Explanation:
The equivalent mass of
Let the amount of
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
And 30.10mL 0.1220M or N NaOH solution will contain
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