Question

Consider the titration of a 25.0 −mL sample of 0.110 M HC2H3O2 with 0.125 M NaOH. Determine each of the following.(please do all)

a)the initial pH

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

c)the pH at 6.00 mL of added base

d)the pH at one-half of the equivalence point ,andt he pH at the equivalence point

Answer 1

a)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.11 0 0

0.11-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.11) = 1.407*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

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

1.98*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.98*10^-6 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.8*10^-5

c = -1.98*10^-6

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 7.92*10^-6

roots are :

x = 1.398*10^-3 and x = -1.416*10^-3

since x can't be negative, the possible value of x is

x = 1.398*10^-3

use:

pH = -log [H+]

= -log (1.398*10^-3)

= 2.8544

Answer: 2.85

b)

use:

pKa = -log Ka

4.818 = -log Ka

Ka = 1.521*10^-5

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.11 M *25.0 mL = 0.125M *V(NaOH)

V(NaOH) = 22 mL

Answer: 22 mL

c)

Given:

M(CH3COOH) = 0.11 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.125 M

V(NaOH) = 6 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.11 M * 25 mL = 2.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.125 M * 6 mL = 0.75 mmol

We have:

mol(CH3COOH) = 2.75 mmol

mol(NaOH) = 0.75 mmol

0.75 mmol of both will react

excess CH3COOH remaining = 2 mmol

Volume of Solution = 25 + 6 = 31 mL

[CH3COOH] = 2 mmol/31 mL = 0.0645M

[CH3COO-] = 0.75/31 = 0.0242M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {2.419*10^-2/6.452*10^-2}

= 4.319

Answer: 4.32

d)

i) At half equivalence point,

pH = pKa

- log (Ka)

= - log(1.8*10^-5)

= 4.745

Answer: 4.75

ii) At equivalence point

Given:

M(CH3COOH) = 0.11 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.125 M

V(NaOH) = 22 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.11 M * 25 mL = 2.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.125 M * 22 mL = 2.75 mmol

We have:

mol(CH3COOH) = 2.75 mmol

mol(NaOH) = 2.75 mmol

2.75 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 2.75 mmol

Volume of Solution = 25 + 22 = 47 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.521*10^-5 = 6.577*10^-10

concentration ofCH3COO-,c = 2.75 mmol/47 mL = 0.0585M

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.0585 0 0

0.0585-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((6.577*10^-10)*5.851*10^-2) = 6.203*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.203*10^-6 M

[OH-] = x = 6.203*10^-6 M

use:

pOH = -log [OH-]

= -log (6.203*10^-6)

= 5.2074

use:

PH = 14 - pOH

= 14 - 5.2074

= 8.7926

Answer: 8.79