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The resulting pH due to the mixing of four solutions

  1. Mar 23, 2009 #1
    ive spent over 2 hours on this question and i am unable to solve it. If anyone could help it would be greatly appreciated.

    What is the pH of the solution that results from mixing the following four solutions together?

    (1) 150. mL of 0.250 M NaCl
    (2) 300. mL of 0.200 M HCl
    (3) 100. mL of 0.050 M HNO3
    (4) 450. mL of 0.200 M NaOAc

    HCl and HNO3 are both strong acids and give 0.065 mol of h3o.
    There are also 0.09 mol of NaOAc which is a weak base in the solution.
    The hydronium reacts with NaOAc, resulting in 0.025 mol of NaOAc in a 1L solution

    i do not know what to do after, ive reacted so many possible combinations and still cannot get an answer. The possiable answers are as follows:
    a. ) 4.22
    b. ) 4.33
    c. ) 4.74
    d. ) 4.98
    e. ) 5.16
     
  2. jcsd
  3. Mar 23, 2009 #2

    symbolipoint

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    edit: I misunderstood the question when I posted this response:

    The easier situations are #1, 2, 3. #1 is only a neutral salt in solution; #2 and #3 are each strong acid situations (therefore, completed ionized in solution). Use the definition of pH (or the simpler definition of pH), being "negative logarithm, base 10, of the hydrogen ion concentration as molarity".

    #4 is a bit more complicated in that you need to be studying weak acid equilibrium.
     
    Last edited: Mar 23, 2009
  4. Mar 23, 2009 #3
    that doesnt really help me lol, im really stuck
     
  5. Mar 23, 2009 #4

    symbolipoint

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    Right. I misunderstood your question. You wanted to know the pH when the FOUR listed solutions are mixed. This will be a weak-acid equilibrium problem. The salt solution will have no effect. The solutions of strong acids will contribute hydronium ions. This will have the effect of inhibiting ionization of the acetic acid. There is a good working expression for how to handle this using formal concentrations of acetic acid and any (if any) formal salt concentration of the acetic acid. Unfortunately, I am not in current condition to handle that expression (I have been in the past). I may need to review before I offer the help; meanwhile, someone else may give this help before I do my review of this topic.

    From me, maybe two or three hours to restudy - that's why someone else will probably help you before I do.
     
  6. Mar 23, 2009 #5

    symbolipoint

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    I thought about this a bit longer, and you can assume that most or nearly all of the acidity comes from the strong acids, so that ionization from acetic acid is negligible. That should simplify the exercise. Hydroxide will be unimportant in the exercise, also.

    Fundamentally you can use (H+Ac-OH)(H)/(Fsalt-H) = Ka

    but practically, you may just as well use (H+Ac)(H)/(Fsalt-H) = Ka or also better
    (H)(H)/(Fsalt-H) = Ka because you could expect negligible dissociation of acetic acid.

    [tex]\frac{(H)(H)}{Fsalt - H}[/tex] = Ka in case the pure text formatting be difficult to read.
     
  7. Mar 24, 2009 #6

    Borek

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    This is simple buffer problem, just convoluted so that it is hard to spot :smile: Assume protonation went to completion, use Henderson-Hasselbalch equation.
     
  8. Mar 24, 2009 #7

    symbolipoint

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    Another approach is to check for any excess of hydrogen ions after protonating the acetate ion. Any acidic pH is essentially all from the unused hydrogens from the strong acids.
     
  9. Mar 24, 2009 #8

    Borek

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    Yes, but there is no excess H+.
     
  10. Mar 24, 2009 #9

    symbolipoint

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    Very good. I did not check that far. My statement was, at least in part, a qualitative best guess. I really did not do the mole for mole calculations.
     
  11. Mar 24, 2009 #10

    Borek

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    To be honest I was too lazy to do it by hand, but my pH calculator (well, one of my pH calculators, the more universal one) makes wonders in such cases.
     
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