For the diprotic titration you will be performing in the lab your unknown soluti
ID: 488566 • Letter: F
Question
For the diprotic titration you will be performing in the lab your unknown solution will be a mixture of PO43– and HPO42–. At the first equivalence point all of the PO43- will have been converted to HPO42-, so the solution will now contain only HPO42– . In a similar calculation to that in the previous question, a student has found that their original solution contained 1.409 mmol of PO43-. If reaching the second equivalence point consumed 20.24 mL more of 0.1384 M HCl, how many mmol of HPO42– were in the original sample? Report answer to 3 decimal places. No units necessary.
Explanation / Answer
moles of HCl = 0.02024 x 0.1384
= 2.80 x 10^-3 mol
= 2.80 mmol
millimoles of PO43- = 1.409 mmol
Since 1.409 mmol of that was initially PO43-, then
mmol of HPO42- initially present = 2.80 - 1.409
mmol of HPO42– were in the original sample = 1.392 mmol