Calculate the pH during the titration of 30.0 ml of 0.40M HCl with a 0.20M NaOH
ID: 1053465 • Letter: C
Question
Calculate the pH during the titration of 30.0 ml of 0.40M HCl with a 0.20M NaOH solution at the following point during the titration.
a) 0.00ml of NaOH added
millimoles acid initially=__________
Millimoles base added= ___________
pH = ____________
b) 20.0 ml of NaOH added
millimoles acid initially=__________
Millimoles base added = ___________
pH = ____________
c)60.0ml NaOH added
millimoles acid initially=__________
Millimoles base added = ___________
pH = ____________
d) 65.00ml NaOH added
millimoles acid initially=__________
Millimoles base added = ___________
pH = ____________
Explanation / Answer
a) millimoles of acid initially = 30 x 0.4 = 12
millimoles of base added = 0
pH = -log [H+]
pH = -log [0.4]
pH = 0.398
b) millimoles of acid initially = 30 x 0.4 = 12
millimoles of NaOH added = 20 x 0.2 = 4
[HCl] = 8 / 50 = 0.16 M
pH = - log [0.16]
pH = 0.796
c) millimoles of acid = 30 x 0.4 = 12
millimoles of NaOH = 60 x 0.2 = 12
equivalence point
pH = 7.0
d) millimoles of acid = 30 x 0.4 = 12
millimoles of NaOH added = 65 x 0.2 = 13
13 - 12 = 1
[NaOH] = 1 / 95 = 0.010 M
pOH = - log [0.01]
poH = 2.0
pH = 14 - 2.0
pH = 12.0