Question

8. An analytical chemist is titrating 79.9 mL of a 0.7300 M solution of formic acid (HCOOH) with a 0.6000 M solution of KOH. The pk, of formic acid is 3.74 Calculate the pH of the acid solution after the chemist has added 19.5 ml of the KOH solution to it. Note for advanced students you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round you answer to 2 decimal places

Answer 1

pH after addition of 19.5 mL KOH = 3.14

Moles of KOH added = 0.6000 mot/L x 0.0195L = 0:0117 mo1 Reaction: H2O HCOOH +OH T:l mola ratio, KOH reacts en As HCoo H and HCOOH neutralized = 0.0117 mol of motes x 0.0799L HCooH = 0.1300 mol/L Initàl moles = 0:058327 mol HCOOH = HCOO Inital: 0.0s8327mol O. 0117 mol O. O117 mol Change : -0·0117 mol O. 0117 mol : 0:046627 mol 0-0117 mol Equilibrium : han Hende son - Hasselbach equatcon, pH = pKa + log ((A-]) (6HA], %3D uolume is same for both A and HA, use the modes in place Since of (AS and (HAJ. can we Now, (A] PH = Pka + log [HAJ O 0117 log 3.74 + 0.0 46627 (-0.600) = 3.74 + 3.14