Consider the titration of 25.0 mL of 0.125M HCl with 0.100M KOH. Calculate the pH?

HCl + KOH --------> KCl + H2O

no.of moles of HCl = molarity X volume in L = 0.125 X 25/1000 = 0.00313

when you add 3 ml of KOH .....no.of moles of KOH present in it = 0.1 X 3/1000 = 0.0003

as it is a 1:1 reaction ....so 0.0003 moles of KOH will react with 0.0003 moles of HCl

so no.of moles of KOH left = 0

no.of moles of HCl left = 0.00313 - 0.0003 = 0.00283

no need to calculate how much KCl is being formed as KCl will not have any effect on pH as it is a salt of strong acid and strong base

final volume = 25 + 3 = 28 ml = 0.028 ml

new molarity of HCl = 0.00283/0.028 = 0.101

as HCl is a strong acid ....so it dissociates completely

HCl --------> H+ + Cl-

so [H+] = [HCl] = 0.101 M

pH = -log [H+] = -log 0.101 = 0.996

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now when you add 20 ml of KOH ...no.of moles = 0.1 X 20/1000 = 0.002

no.of moles of HCl left = 0.00313 - 0.002 = 0.00113

total volume = 20 + 25 = 45 ml = 0.045 L

new molarity = 0.00113/0.045 = 0.0251 M

[H+] = 0.0251 M

pH = -log [H+] = -log 0.0251 = 1.6