[DMOC]

Question: Citric acid (H3C6H5O7) has 3 acid hydrogens. It takes 21.28 mL of 0.250 M NaOH to titrate 10.0 mL of the citric acid. What is the mass of citric acid that was dissolved in the 10.00 mL sample of the acid solution?

I did this problem by using M₁V₁ = M₂V₂

V1 = 0.010 L
M2 = 0.250 M
V2 = 0.02128

To get M1 = 0.00532 M of citric acid.

Then I found moles of citric acid and converted that to grams, to get 1.02 g of citric acid.

My teacher is telling me, though, that I should avoid that formula for titration.


How else do  isolve this problem?

[Arctic-Nation]

While the formula you used is valid when working with monoprotic acids and bases, it is more difficult to use it for polyprotic acids and bases. A more general approach would be to calculate the number of moles of NaOH, then convert it to the number of moles of acid. In your example, 1 mole of NaOH reacts with only 1/3 mole of citric acid. Once you have that, it's easy to calculate mass and concentration.

[DMOC]

Er, what do I do once I have the moles of citric acid?

[Borek]

Are you asking how to convert number of moles to mass?

[CIE alevel]

Er, what do I do once I have the moles of citric acid?

Er, what u did earlier perhaps?? lol  =D

As arctic - nation says....u find the moles of citric then calc. mass...
Basically the mass u must have calculated after using formula for titration.

your teacher must be saying to do it step-wise rather than using M1V1 = M2V2 !
Perhaps u missed out the stoichiometry of the acid-base reaction, it happens with that formula!!