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?

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1. 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.

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posted by DrBob222
2. 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.

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posted by Kat
3. Buffers increase ph of acids, they buffer the degree at which acids affect solutions

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posted by Cheggy

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