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 = ____________

Answer 1

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