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Calculate [H3O] in the following solutions.
5.15x10-2M HCl and 7.62x10-2M NaC2H3O2


HCl + NaC2H3O2. It appears there is an excess of H^+ which will react with NaC2H3O2 to form HC2H3O2. That gives you a buffer of C2H3O2 and HC2H3O2 and you use the H-H equation to solve for pH.
Post your work if you get stuck.

i used the H-H equation.
pKa=4.74, conjugate base= 7.62x10-2, Acid=5.15X10-2. i got a pH 0f 4.91 and H30 turned out to be 1.2X10-5.
Is this right?

  • chem -

    I don't think so.
    Now that I see the question again, I'm wondering if this is a problem with two parts; i.e., the question is asking for the pH of 0.0515 M HCl and for the pH of a separate solution of 0.0762 M NaC2H3O2? And after noticing that there is no volumes listed for EITHER solution, I suspect this may be the right interpretation. So the first would be the pH of HCl and the second would be the pH of a sodium acetate solution and my first response would be null and void.
    IF, however, it really is a mixture of two solutions, I started with 100 mL of each. That gives me 5.15 mmoles HCl and 7.62 mmoles NaC2H3O2. The final product is 5.15 mmoles (which you have correctly identified) BUT the sodium acetate remaining is 7.62 mmoles - 5.15 mmoles = 2.47 mmoles.
    pH = 4.74 + log(2.47/5.15) = 4.42. Check my work but especially look at the question again. I'll bet you are to determine the pH of EACH solution separately and it isn't a mixture.

  • chem -

    i'm problems with the same question, did you find out the correct answer?
    and to clarify, the question is asking the [H30+] concentration when the 2 are mixed, so you have to use the common ion effect method.

  • chem -

    Calculate [H3O] in the following solutions.
    5.15x10-2M HCl and 7.62x10-2M NaC2H3O2

    [H3O+] = 4.0×10−5 M

  • chem -

    I tried 4.0x10^-5 with a slightly different first value (5.35E-2 vs 5.15E-2) and it was still marked as incorrect. How do i get the correct answer

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