Question

The next 11 questions are related to the titration of 20.00 mL of a 0.0950 M acetic acid solution with 0.0700 M KOH. Assume that the temperature is 25 ^oC.
What is the initial pH of the analyte solution?

What is the pH when 17.00 mL of the KOH solution have been added?

What is the pH when 33.00 mL of the KOH solution have been added?

Answer 1

1)when 0.0 mL of KOH is added

CH3COOH dissociates as:

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

9.5*10^-2 0 0

9.5*10^-2-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)*9.5*10^-2) = 1.308*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/(9.5*10^-2-x)

1.71*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.71*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.71*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 = 6.84*10^-6

roots are :

x = 1.299*10^-3 and x = -1.317*10^-3

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

x = 1.299*10^-3

use:

pH = -log [H+]

= -log (1.299*10^-3)

= 2.8865

Answer: 2.89

2)when 17.0 mL of KOH is added

Given:

M(CH3COOH) = 0.095 M

V(CH3COOH) = 20 mL

M(KOH) = 0.07 M

V(KOH) = 17 mL

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

mol(CH3COOH) = 0.095 M * 20 mL = 1.9 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.07 M * 17 mL = 1.19 mmol

We have:

mol(CH3COOH) = 1.9 mmol

mol(KOH) = 1.19 mmol

1.19 mmol of both will react

excess CH3COOH remaining = 0.71 mmol

Volume of Solution = 20 + 17 = 37 mL

[CH3COOH] = 0.71 mmol/37 mL = 0.0192M

[CH3COO-] = 1.19/37 = 0.0322M

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 {3.216*10^-2/1.919*10^-2}

= 4.969

Answer: 4.97

3)when 33.0 mL of KOH is added

Given:

M(CH3COOH) = 0.095 M

V(CH3COOH) = 20 mL

M(KOH) = 0.07 M

V(KOH) = 33 mL

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

mol(CH3COOH) = 0.095 M * 20 mL = 1.9 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.07 M * 33 mL = 2.31 mmol

We have:

mol(CH3COOH) = 1.9 mmol

mol(KOH) = 2.31 mmol

1.9 mmol of both will react

excess KOH remaining = 0.41 mmol

Volume of Solution = 20 + 33 = 53 mL

[OH-] = 0.41 mmol/53 mL = 0.0077 M

use:

pOH = -log [OH-]

= -log (7.736*10^-3)

= 2.1115

use:

PH = 14 - pOH

= 14 - 2.1115

= 11.8885

Answer: 11.89