Phosphoric acid titration calculation

[tq1088]

Homework Statement

The problem is that i am trying to find out the concentration of phosphoric acid in cola. I have completed the titration of 3 different types of cola. I have obtained the pH and volume on the titration graph. Note: I am trying to figure out the overall concentration of H3P04 first, not the H+ ions. I will do that later, but first i need to find the concentration of the H3P04. I have results like:

1st equivalence point: pH is 4.87 and Volume is 2.607

When 2.607ml of sodium hydroxide was added, the pH at the first equivalence point is 4.87.
The same follows for the second equivalence point.

2nd equivalence point: pH is 9.27 and Volume is 6.983.

Btw, the concentration of the sodium hydroxide is 0.1M and the amount of cola used is 50ml

The Attempt at a Solution

Note: The below example is with different values...from above

I have already done some calculating where i used a 3:1 ratio to find the concentration of the 1st and 2nd equiv point, but i think these are incorrect. For example i did this:

NaOH --> Volume is 1.964 (0.001964L) and Concentration is 0.1M H3P04 --> Volume = 50ml (0.05L) and concentration is unknown.

Ratio is 3:1 So 0.001964 x (1/3) = 6.54 X 10^-4

So H3P04 concentration is 6.54 x 10^-4 / 0.05

= 0.013M

I did the same for the 2nd equiv point but i used 3.286ml (0.003286ml) instead and i got 0.022M

I don't think this is correct because i need to find the overall concentration of the phosphoric acid.This is just one example of mine. I have much more results. If someone could help me calculate the concentration of the acid in the cola, i would really appreciate it.

Thanks

Homework Equations

This is the overall equation.

3NaOH + H3PO4 ->Na3PO4 +3H2O

 

[Borek]

What has been neutralized at the first equivalence point? Write reaction equation - is it 3:1?

Do the same for the second equivalence point.

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[tq1088]

1st equiv point

H3PO4 (aq) --> H+ (aq) + H2PO4−

2nd equiv point

H2PO4 (aq) --> H+ (aq) + H2PO4 2−

3rd equiv point

HPO4 2− (aq) --> H+ (aq) + PO43−

But the 3rd doesn't really matter because it didn't show up on the titration graph.

The overall equation is

H3PO4 (aq) + 3NaOH (aq) --> Na3PO4 (aq) + H2O (l)

3:1 because the reaction occurs between 3 of NaOH and 1 of H3P04

and I am trying to find the overall concentration of the H3P04 and i got 2 equiv points on the titration graph.

 

[tq1088]

Like the amount of NaOH needed to reach the 1st equiv point is 1.964ml. The pH was read at 5.14Then to reach the 2nd equiv point, 3.286 ml of NaOH is needed. The pH was read at 9.17

 

[Borek]

Geez, please read what I wrote and answer my question. I am not asking about overall reaction equation, I am asking what have been neutralized at the first equivalence point.

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