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Finding salt concentrations :

  1. Sep 23, 2004 #1
    can someone please help me with this problem, its giving me lots of trouble to solve :

    (a) Suppose you wanted to make a buffer of exactly ph 7.00 using KH2PO4 and Na2HPO4. If you had a solution of 0.1M KH2PO4, what concentration of Na2HPO4 would you need?

    (b)Now assume you wish to make a buffer at the same ph, using the same substances, but want the total phosphate molarity (HPO42-) + (H2PO4-) to equal 0.3 What concentrations of KH2PO4 and Na2HPO4 would you use?
     
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  3. Sep 23, 2004 #2

    Gokul43201

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    What have you tried so far ? What ideas do you have ?

    Tell us what you've tried and we'll guide you the rest of the way. :smile:
     
  4. Sep 23, 2004 #3
    i tried the henderson/hasselbalch equation :
    the salt : Na2hpo4
    the acid :kh2po4
    ph= 7.0
    pka ? not sure
    not sure how to proceed on the second part or how to finish the first part
     
  5. Sep 24, 2004 #4

    chem_tr

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    If you want to find pH, then you should use pKa of the acid in Henderson-Hasselbach equation.
    In your second question, you don't know the molar ratio of conjugate base to acid. The rest is known, just find it.
     
  6. Sep 29, 2004 #5
    question

    its been said that the pka of strong bases are so close to 1 they ( the pka ) can be ignored.

    Is this true ?
     
  7. Sep 29, 2004 #6

    chem_tr

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    Yes that's correct, but it is invalid for this question, as phosphate species other than ortho-phosphoric acid (H3PO4) need pKa values. If you were not given that data, it is nearly not possible to solve the H-H equation. Please find the pKa values for them, and solve this question. It is not that difficult. But when you think you're lost, keep asking.
     
  8. Sep 29, 2004 #7
    thanks

    i have already solved the problem i was just asking a general question
    thanks chem_tr
     
  9. Oct 1, 2004 #8
    new question

    I was given this lab problem calc the ph values and draw the titration curve for the titration of 500ml of 0.010m acetic acid (pka 4.76) with .010m KOH.
    I have to give two points the starting ph and the equivalence point.

    I found the equivalence point to be the pka ( where ph=pka ).
    I know i have to calc the second point using the KA Im just not sure how to plot the x-axis and y-axis

    for the y-axis i know i have to plot ph
    but for the x-axis is it KOH (eq ) or what ?
    im not sure
    can someone help me with this one ?
     
  10. Oct 3, 2004 #9

    chem_tr

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    Well, the starting pH will be the pH of the acetic acid solution with no KOH added, I am sure you are not asking this. What you're asking instead is simple, too; just plot the curve between titer (mL of KOH solution) and pH, on x and y axis, respectively.
     
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