Question

titration of 25.00 ml of 0.100M CH3COOH with 0.100 M NaOH (weak acid, strong base)

a) calculate the initial pH ( kb = 1.8 x10 ^-5)

b) why is pH > 7 at the equivalence point?

c) calculate the pH at the equivalence point

Answer 1

a) Initial pH

Set up ICE using dissociation reaction of acetic acid.

            CH3COOH (aq) + H2O (l) --- > CH3COO-(aq) + H3O+(aq)

I           0.100                                           0                         0

C          -x                                                  +x                     +x

E        (0.100-x)                                         x                            x

Use ka expression to find x

Ka = 1.8E-5 = x² / ( 0.100-x)

X in the denominator is neglected since the value of ka is small

1.8E-5 = x²/ 0.100

x = 0.001342 = [H3O+]

pH = - log ( 0.001342)

= 2.87

b)

Write the reaction between NaOH and acetic acid

CH3COOH (aq) + NaOH (aq) -- > CH3COONa (aq) + H2O (l)

At equivalence point all the acetic acid moles are neutralized by NaOH and therefore there is only CH3COONa in the solution which is basic salt and that produces OH- ions. When OH- are produced in the solution, it become basic and the pH of the solution would be less than 7

C)

pH at equivalence point

moles of CH3COONa = moles of CH3COO-       …( salt dissociates completely)

molarity of CH3COO- = mole/ total volume

total volume = volume of acetic acid + volume of base

Moles of acetic acid = volume in L x molarity = 0.025 L x 0.100 = 0.0025 mol

Moles of NaOH = 0.0025

Volume of NaOH = 0.0025 mol / molarity = 0.0025 mol / 0.100 = 0.025 L = 25.0 mL

Therefore total volume = 50 mL = 0.050

Molarity of CH3COO- = 0.0025 / 0.050 = 0.05 M

Set ICE

            CH3COO-(aq) + H2O (l) ---- > CH3COOH (aq) + OH- (aq)

I       0.05                                                 0                               0

C       -x                                                   +x                            + x

E    (0.05-x)                                             x                                 x

Calculate kb using ka

Kb = 1.0E-14 / 1.8E-5 = 5.55 E-10

Kb = x² / (0.05-x) = 5.55E-10

Since the value of kb is very small we can neglect x in the denominator.

x^2 = 5.55E-10 x 0.05

x = 5.26 E -6 = [OH-]

pOH = - log [OH-] = - log ( 5.26 E-6) =5.28

pH = 14- pOH = 14 – 5.28 = 8.72

pH = 8.72