Question

calculate the pH during the titration of 50.00 ml of .400 M HCl with .800 M KOH after 0, 12.50, 25.00, and 40.00 ml of KOH have been added. Graph the titration curve.

Answer 1

HCl is strong acid dissociates completly to give H+

KOH is strong base and gives completely to give OH-

1) Initially 50 ml of 0.4 M HCl present when 0 ml KOH added , [H+] = [HCl] = 0.4 M

pH = -log [H+] = -log ( 0.4) = 0.4

2) HCl moles = M x V = 0.4 x ( 50/1000) = 0.02

KOH moles added = 0.8 x 12.5/1000 = 0.01

HCl moles left after neutralisation = 0.02-0.01 = 0.01

total vol of solution = 50+12.5 = 62.5 ml = 62.5/1000 L = 0.0625 L

[H+] = ( 0.01/0.0625) = 0.16

pH = -log ( 0.16) = 0.8

3) when 25 ml KOH added , KOH moles = M x V = 0.8 x ( 25/1000) = 0.02

HCl moles = KOH moles , hence after neutralisation we have neutral pH

pH = 7

4) KOH moles = ( 0.8 x 40/1000) = 0.032

excess KOH moles after reacting with 0.02 moles HCl = 0.032-0.02 = 0.012

total vol = 50+40 = 90 ml = 0.09 L

[OH-] = ( 0.012/0.09) = 0.13

pOH = -log ( 0.133) = 0.875 , pH = 13.12

Plot of pH vs vol of KOH added is done