Calculate the pH in a buffer prepared from $\pu{50 mL, 0.30 M}$ formic acid ($\ce{HCOOH}$) and $\pu{30 mL, 0.40 M}$ sodium formate ($\ce{HCOONa}$).

My way of solving:

$n_{\ce{HCOOH}} = 0.05 \times 0.3 = \pu{0.015 mol}$

$n_{\ce{HCOONa}} = 0.03 \times 0.4 = \pu{0.012 mol}$ (which means $\ce{NaOH} = \pu{0.012mol}$)

$\ce{HCOOH + NaOH <=> HCOONa + H2O}$

After doing a BCA-table, I got that the remaining moles of HCOOH = $\pu{0.003 mol}$

and HCOONa = $\pu{0.012 mol}$

Calculating the new concentration:

[HCOOH] = $\ce{0.003 / 0.08 = 0.0375M}$

[HCOONa] = $\ce{0.012 / 0.08 = 0.15M}$

Using the Henderson–Hasselbalch equation:

pH = pKa + log [HCOONa]/[HCOOH] = $\ce{3.75 + 0.60 = 4.35}$

So my answer is that the pH of the buffer is $\ce{4.35}$

Answer key

However, in our answer key, they found the new concentration of HCOOH and HCOONa by simply taking:

[HCOOH] = $\ce{(0.05 *0.3)/0.08 = 0.1875M}$

[HCOONa] = $\ce{(0.03 *0.4)/0.08 = 0.15M}$

and when using those concentrations in the Henderson–Hasselbalch equation they get the answer pH = $\ce{3.65}$


Is the answer key correct? If we are adding a base, then the pH should increase to become more basic, not decrease as in the answer key. If the answer key is correct, why does the pH decrease?

  • 1
    $\begingroup$ The answer key's method is certainly valid given that the system is buffered. I suspect that your calculations for final concentrations are incorrect. For one thing, you should be doing those calculations on concentrations, since the equilibrium expression all use concentrations, not amount of substance. $\endgroup$
    – Zhe
    Mar 8, 2019 at 16:35
  • $\begingroup$ @Zhe But shouldn't the pH increase and become more basic? $\endgroup$
    – katara
    Mar 8, 2019 at 17:06
  • 1
    $\begingroup$ @katara - No look at your calculation of the moles. There are 0.015 moles of formic acid and only 0.012 moles of sodium formate. Thus you would expect the solution to be more acidic than the pKa value. $\endgroup$
    – MaxW
    Mar 8, 2019 at 17:54
  • $\begingroup$ @katara - you shot yourself in the foot by going off on a tangent and thinking about NaOH. It is true that if you started with 0.027 moles of formic acid and 0.012 moles of NaOH then you would end up with 0.015 moles of formic acid and 0.012 moles of sodium formate, but why go "backwards" to solve the problem? $\endgroup$
    – MaxW
    Mar 8, 2019 at 18:43

1 Answer 1


Is the answer key correct?

Yes, they have a very good approximation of the pH, and their way of solving it is straightforward and conventional.

After doing a BCA-table, I got that the remaining moles of HCOOH = 0.003moles

The sum of formic acid and formate has to remain 0.027 mol. You could start out with 0.027 mol of formic acid and turn some of it into formate by adding NaOH. Or you could just mix them as described in the problem. However, you are claiming that there is a total of 0.015 mol of formic acid plus formate, so you lost some atoms there. As the comments state, there is no reason to think about where they got the sodium formate (stock room?).

If the answer key is correct, why does the pH decrease?

Decrease or increase depends on your frame of reference. If you compare it with a 1:1 buffer where the pH = pKa, it decreased because there is a higher concentration of acid than conjugate base.

If you compare it to the pH of 0.3 M formic acid (roughly pH = 2.1), then it increased because you added conjugate base (neutralized and diluted the acid).


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