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
For me the answer is E. It is composed of one Ru2+ ion and two F- ions.
