A 2.5gm sample of a mixture od sodium carbonate and sodium chloride is dissolved in 25ml 0.798M HCl. Some acid remains after the treatment of the sample.

If 28.7ml of 0.108M NaOH were required to titrate the excess HCl how many moles of sodium carbonate were present in the original sample?

--> I have used N = C x V to work out the number of moles of HCl (19.95), NaOH (3.10) & how many moles of HCl were titrated by the 2.5g mixture (16.85),
however I am now confused on how to work it out for sodium carbonate?

please help

Quote:

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Originally Posted by tulip88

A 2.5gm sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25ml 0.798M HCl. Some acid remains after the treatment of the sample.

If 28.7ml of 0.108M NaOH were required to titrate the excess HCl how many moles of sodium carbonate were present in the original sample?

--> I have used N = C x V to work out the number of moles of HCl (19.95), NaOH (3.10) & how many moles of HCl were titrated by the 2.5g mixture (16.85),
however I am now confused on how to work it out for sodium carbonate?

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The first equation you need to understand is:



The second equation is



{note in the following that 1 molar concentration is 1 mole / liter or 1 mmole / milliliter. A millimole (mmole) is a thousanth of a mole. This unit can sometimes be more convenient than working in moles.}

You added
25 mL x 0.798 mmoles/mL = 19.95 mmoles of HCl
and "back titrated" with
28.7 mL x 0.108 mmoles/mL NaOH = 3.0996 mmoles NaOH

Since one mole of HCl reacts with one mole of NaOH (see the second equation), you can subtract the two to get the number of moles of HCl that would have been required to completely react with the sodium carbonate (Na2CO3).

19.95 - 3.0996 = 16.8504 moles of HCl used

From the first equation, you can see that 2 moles of HCl react with one mole of Na2CO3. Therefore, there must have been

16.8504 / 2 = 8.4252 mmoles of Na2CO3

or

Thank you so much! That makes so much more sense now.
Really appreciated!