Post a New Question

Chemistry lab

posted by .

1) Using the 1st equivalence point from part a calculate the molarity of the phosphoric acid. Molarity of NaOH = .1523 M
Volume of phosphoric acid used 10.0mL
Volume at equivalence point 7.410 mL and 14.861 mL

2) Using the 2nd equivalence point from part a, calculate the molarity of the phosphoric acid. (remember that 2 H+ ions are being neutralized in this case.) How does this value compare to question one?
Any help would be appreciated! Thanks!

  • Chemistry lab -

    NaOH + H3PO4 ==> NaH2PO4 + H2O
    mL x M NaOH = millimols.
    That = mmols H3PO4
    M H3PO4 = mmols H3PO4/mL H3PO4 = ?

    2NaOH + H3PO4 ==>Na2HPO4
    mL x M NaOH = 14.861 x 0.1538 = ?
    mmols H3PO4 = ? x 1/2
    M H3PO4 = mmols H3PO4/10 mL = ?

  • Chemistry lab -

    Thank you for your response! Having all of these numbers confused me. So for part a I'm multiplying 7.410 mL (first equivalence point) by .1523 M. I got 1.129 mmol H3PO4. Then divided it by 10.0mL to get a concentration of 0.113 M H3PO4.

  • Chemistry lab -

    I see I made a typo on molarity. The 0.1538 I used should be 0.1523.
    You're right. 7.410 x 0.1523/10.0 = ?
    14.810 x 0.1523 x 1/2 x 1/10.0 = ?

  • Chemistry lab -

    Thanks so much and the typo was no big deal I knew what you meant. :) you're a life saver!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question