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Analytical Chemistry

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Phosphoric acid is a triprotic acid with the following pKa values:

pka1: 2.148 pka2: 7.198 pka3: 12.375

You wish to prepare 1.000 L of a 0.0100 M phosphate buffer at pH 7.45. To do this, you choose to use mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?

?? grams NaH2PO4
?? grams Na2HPO4

Hint: Use the Henderson-Hasselbalch equation to get the molar ratio of Na2HPO4 to NaH2PO4 required, then the fraction of each form from the ratio. The total moles needed will be 1.000 L × 0.0100 M = 0.0100 moles. Use the formula mass to calculate the mass needed. (FM NaH2PO4 = 119.98; FM Na2HPO4 = 141.96).

I am completely lost on how to solve this problem.

  • Analytical Chemistry -

    pH=pka+log({A-]/[HA])


    pH=7.45
    pka=12.375


    Solve for ratio:

    10^(7.45-12.375)={A-]/[HA]=1.19x10^-5

    Since you have a total of 0.01 moles,

    1.19x10^-5*(0.01moles)= moles of Na2HPO4

    0.01 moles-moles of Na2HPO4= moles of NaH2PO4


    Use the formula weights to solve for the number for each that Dr. Bob222 gave you.


    **** Not sure about that pKa value that I chose.

  • Analytical Chemistry -

    The correct pKa value to choose is pK2.

    Use the HH equation to solve for the ratio base/acid.
    One equation you need is base/acid.
    The other equation you need is
    base + acid = 0.01
    That two equations in two unknowns; solve for acid concn and base concn and convert to grams. Post your work if you gets stuck.

  • Analytical Chemistry -

    Setup is correct, but substitute 7.198. That should change 1.19x10^-5 to 1.78 and do what Dr.Bob222 told you to do.

  • Analytical Chemistry -

    if you use the pk2 wouldnt you get a negative value for nah2po4?

  • Analytical Chemistry -

    After you get the ratio. From the
    0.01 moles-moles of Na2HPO4= moles of NaH2PO4
    => (0.01 - Na2HPO4)/(Na2HPO4) = (ratio of pka2 from hasselbalch)
    unit will be mol, convert to g

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