Question

A buffer is 0.282 m c6h5cooh(aq and 0.282 m na(c6h5coo(aq. calculate the ph after the addition of 0.150 moles of nitric acid to 1.0 l of the buffer. the pka of c6h5cooh is 4.20.

Answer 1

You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:

PH = Pka + log[acid/base]

Step 1 (Figure out the concentrations):

0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3

Step 2 (Plug into equation):

PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]

PH = 3.69