Question

Consider the titration of a 21.0 mL sample of 0.110 molL−1 CH3COOH (Ka=1.8×10−5) with 0.125 molL−1 NaOH. Determine each quantity:

a) the initial pH. Express your answer using two decimal places.

b) the volume of added base required to reach the equivalence point

c) the pH at 4.0 mL of added base. Express your answer using two decimal places.

d) the pH at one-half of the equivalence point. Express your answer using two decimal places.

e) the pH at the equivalence point. Express your answer using two decimal places.

f) the pH after adding 6.00mL of base beyond the equivalence point. Express your answer using two decimal places.

Answer 1

a)

First, assume the acid:

CH3COOH

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.11 M; then

x^2 + (1.8*10^-5)x - 0.11 *(1.8*10^-5) = 0

solve for x

x =0.01434

substitute

[H+] = 0 + 0.01434= 0.01434 M

pH = -log(0.01434) = 1.834

b)

V for equivalence

mmol of acid = MV = 21*0.11 = 2.31

Vbase = mmol/M = 2.31/0.125 = 18.48 mL

c)

mmol of acid =  21*0.11 = 2.31

mmol of base = MV = 0.125*4 ?= 0.5

after reaction

mmol of acid = 2.31-0.5 = 1.81

mmol of acetate = 0.5

pH = 4.75 + log(0.5/1.81) = 4.19

d)

in the 50% point

pH = pKa + log(50/50)

pH= 4.75 + log(1)

pH = 4.75

e

equivalence

Vtotal = 18.48+21 = 39.48 mL

[A-] = 2.31/39.48 = 0.05851 M

expect hydrolysis

Let HA --> CH2OH and A- = CH2O- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

5.55*10^-10 = x*x/(0.05851 -x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =5.698*10^-6

[OH-]  =5.698*10^-6

pOH = -log(OH-) = -log(5.698*10^-6) = 5.24

pH = 14-5.24= 8.76

f)

extra base:

V of base = (18.38+6) = 24.38

Vtotal = 21+24.38 = 45.38

[OH-] = (6)(0.125) / 45.38

[OH-]= 0.016527

pH = 14+ log(0.016527) = 12.22