Post a New Question

CHEM HELP...PLEASE!!

posted by .

A solution of volume 80.0 mL contains 16.0 mmol HCHO2 and 9.00 mmol NaCHO2.
If 1.00 ml of 12 M HCl is added to this, what will be the resulting pH?

  • CHEM HELP...PLEASE!! -

    Use the Henderson-Hasselbalch equation.
    pH = pKa + log (base/acid)
    First, however, you need to make some adjustments to the acid/base because you have added some HCl to it.
    HCOOH = 16.0 mmole initially.
    HCOONa = 9.00 mmol initially. Adding 1.00 mL of 12 M HCl adds 12 x 1 = 12 mmole (is that 12 or 12.0 or 12.00 mmole?) to this base. Therefore, 9.00 + 12.0 = 21.0 mmole base.
    pH = pKa + log[(21.0)/(16.0)] = ??
    If your prof is picky, s/he will not like this BECAUSE (base) and (acid) are concentrations and not mmols. Technically, then, the concn is mmols/mL = 21.0/81.0 for (base) and 16.00/81.0 for (acid) so the equation is
    pH = pKa + log[(21.0/81.0)/(16.0/81.0)] = ??. You will note that the 81.0 mL volume cancels and some profs just don't put it there. I ALWAYS counted off for not including the volume; however, since it ALWAYS cancels, I would allow the student to use V as in
    pH = pKa + log [(21.0/v)/(16.0/v)] = ??. That way the student let me know that a volume went there but it would cancel and never entered into the calculation.

  • CHEM HELP...PLEASE!! -

    Sorry, there's one thing I don't understand. If an acid is being added to a buffer solution, shouldn't the pH decrease? By following your method, the answer I'm getting is greater than the original pH.

  • CHEM HELP...PLEASE!! -

    Buffers increase ph of acids, they buffer the degree at which acids affect solutions

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question