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 = $\ce{3.65}$
Question
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?