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College Chemistry

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How many milliliters of 1.55 M NaOH must be added to 125 mL of .19 M NaH2PO4 to make a buffer solution with a pH of 6.90?

I have attempted to solve this with no success, as this is an online homework assignment that requires an exact answer.
Hoping that you will be able to help. Thanks.

  • College Chemistry - ,

    80.6 milliliters? O.o

  • College Chemistry - ,

    I'm sorry but this answer didn't work...

  • College Chemistry - ,

    ......H2PO4^- + OH^- ==> HPO4^-2 + H2O
    start..23.75 .....0.........0.........0
    Substitute this into the (B/A) = 0.5036 equation below.

    [mmoles H2PO4 at start is from 125 mL x 0.19 M = 23.75 mmoles.
    x = mmoles OH that must be added.]

    What is the ratio of base to acid?
    6.90 = pk2 + log[(base)/(acid)]
    I looked up k2 for H3PO4 and found 6.34E-8 for pK2 = 7.2 but you need to use the pKa in your work, especially if your data base requires an exact answer. Also, you need to confirm all of my work and round to satisfy appropriate significant figures requirements. Use the Henderson-Hasselbalch equation.
    pH = pKa + log[(base)/(acid)
    6.90 = 7.2 + log(B/A).
    (base)/(acid) = 0.5036 or
    base = 0.5036*acid

    base is x from line 3 above. acid = 23.75-x.
    x = 0.5036(23.75-x)
    Solve for x.
    I get something like 7.95 mmoles for the amount of NaOH to add.

    How much 1.55 M NaOH is needed to give 0.00795 moles. M = moles/L. Solve for L and I get 0.00513 L or 5.13 mL.

    Then you need to check this to make sure it produces a pH of 6.90.
    5.13 mL x 1.55 M NaOH = 7.95 mmoles.
    This will form 7.95 mmoles HPO4^-2(base)
    This will leave 23.75-7.95 = 15.80 mmoles H2PO4^- (acid); then,
    pH = 7.2 + log (7.95/15.8) = 6.90
    Check my thinking. Check my work. Confirm all of this.

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