Question

Consider the titration of 100.0 mL of 0.200 M acetic acid ( Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a 0.0 mL pH= 50.0 mL pH = C 100.0 mL pH = 140.0 mL pH= C 200.0 mL pH = f 240.0 mL pH=

Answer 1

Moles of Acetic Acid- 0.14 x 0.2000 Moles of KOH- - b Ok 0.100 = 0 mol OH- 1 to Subtracting both the Moles- T o lvonta 0.02 molht - 0 mol OH = 0.02 mol Ht dul (excess) Molarity of H+ :- = 0 2 M H+ 0.02 mol Ht Oletgo PH = -log [H + = -log[0.2 pH =0.69 St =1

6 Moles of Acetic Acid = 0.02 mol H+ Moles of KOH 0.05X x 0.100 0.005 mol OH- Subiracting bothe the moles 0.02 mol 4+ - 0.005 mol OH = 0.015 mol H+ (excess) - = 0.1 MHT Molarity of H+ 0.015 mol of Ht 0.1 + 0.05 PH = -log [ht -- log [o.1 [PH = 1 Follow the same method to solve jooc the following other Volume of Koti