Post a New Question


posted by .

Posted by Kat on Monday, March 2, 2009 at 11:21am.

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!! - DrBob222, Monday, March 2, 2009 at 1:25pm

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!! - Kat, Monday, March 2, 2009 at 10:00pm

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.


    Well, you are correct. The pH SHOULD become more acid but it doesn't because I've led you astray. Check my thinking on this.
    The 12 mmoles HCl added uses ALL of the HCOONa and leaves 3 mmoles HCl in excess.
    The 9 mmoles HCOONa has changed to HCOOH and there is now 16 + 9 = 25 mmoles HCOOH (and no HCOONa). So the pH is that of 25 mmoles HCOOH in 81 mL + 3 mmoles HCl in 81 mL. I would use the excess HCl as a common ion to determine the (H^+) contributed by HCOOH to the solution in the presence of HCl, then add it to the (H^+) from the excess HCl. The H^+ contributed by HCOOH may be small enough to neglect but I didn't work it out. In other words, I don't think you have a buffer after the HCl is added.


    This is how I did it:

    HCOOH + H2O <---> HCOO- + H3O+
    I 16.00 9.00
    I 12.00
    C +9.00 -9.00 -9.00
    E 25.00 0 3.00

    HCOOH + H2O <---> HCOO- + H3O+
    I 25.00 0 0
    I 3.00
    C -x +x +x
    E 25 -x x 3+x

    Ka = ------- = 1.8*10^-4

    Assume x is very small, then x = 0.0015

    So, [H30+] = (3+x)mmol / 81ml
    = (3 + 0.0015)mmol / 81 ml
    = 3.0015mol / 81 ml
    = 0.03706 M

    And pH = -log[H30+]
    = 1.43

    I know the technicalities are pretty badly used here, but does my answer make sense?


    yes, it makes sense.
    My answer was similar but slightly different.
    3 mmol H^+ from HCl.
    Ka = 1.77 x 10^-4 (note slightly differnt Ka) = (H^+)(formate)/(HFormate)
    Substitute 3/81 for (H^+) and (25/81 for (Hformate; i.e., formic acid) and solve for formate ion. I get 0.00147 M which is essentially your answer if we used the same Ka value I think.
    Then I converted to moles.
    moles H^+ from formic acid = 0.00147 x 81 mL = 0.119 mmoles.
    moles from HCl = 3 mmoles.
    Add to get 3.119 mmoles
    Then 3.119 mmoles/81 mL = 0.0385 M
    pH = 1.41 which is very close to your answer. Both of us neglected the quadratic if we had solved the Ka with formic acid exactly. The numerator would have been 0.00147 + 0.03704 (we just used 0.03704) and the denominator would have been (0.003086-0.00147)(I used 0.03086 and that could make a difference. A better solution would be to solve that quadratic. In any case, watch the significant figures in the final result.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question