[calvin coolidge]

Do I get "points" for the poetic prose of my subject?  ...anyway, I did a lab where I titrated 30mL of H3PO4 with .103M NaOH and we're to determine the concentration of the H3PO4. The lab manual gave some tips for calculating the acid's concentration: First using the mL of NaOH used to get the 1st equivalence point, then using the mL of NaOH added between the 1st and 2nd equivalence points and average the two concentrations. I found the mL of NaOH used to obtain the 1st equivalence point (14.5mL) from my titration curve. I then calculated the # of moles of NaOH added (.103=n/.0145, n=.0015 moles) to reach this point. The next hint is where I'm stuck. It says to use the stoichiometry of a balanced reaction between NaOH and H3PO4 to determine the # of moles of H3PO4 that I'm titrating and to remember that each step involves the transfer of one proton. So I've bounced around the web searching for examples to help me. I've found one that I originally believed to be right; H3PO4 + 3NaOH-> Na3PO4 + 3H20, giving me an acid:base ratio of 1:3....but then I found a couple more: H3PO4 + NaOH-> NaH2PO4 + H20 and H3PO4 + 2NaOH-> Na2HPO4 + 2H20. Which one(s) are correct? I'm assuming the last two are demonstrating the "proton donation," the 1st equation for the 1st equivalence point and the 2nd for the 2nd equivalence point. Is this right? Sorry for the looooooong description, I didn't want to leave out any pertinent info.

Thanks for any guidance

[Borek]

3NaOH is total neutralization or third equivalence point, 1NaOH is first equivalence point.

[calvin coolidge]

So, am I correct in thinking that, based on the equations H3PO4 + NaOH-> NaH2PO4 + H20 and H3PO4 + 2NaOH-> Na2HPO4 + 2H20, the acid:base mole ratio for the 1st equivalence point is 1:1 and the 2nd equivalence point is 1:2?


Thx,

~CC

[Borek]

Yes.

[calvin coolidge]

...Great!

Thanks for your help, Borek!

~CC