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.
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?
3 answers
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.
Buffers increase ph of acids, they buffer the degree at which acids affect solutions
0 answers