Question

4. 25.0 mL sample of 0.10 M CH COOH is titrated with 0.12 M NaOH. Determine the pH of the solution a) Before the addition of the base. The Ka of CH COOH is 1.8 x 10-5. (10 points) b) After the addition of 15.0 mL of NaOH. The Ka of CH,COOH is 1.8 x 10-5. (10 points)

Answer 1

a)

CH3COOH dissociates as:

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

0.1 0 0

0.1-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)*0.1) = 1.342*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/(0.1-x)

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

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

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

use:

pH = -log [H+]

= -log (1.333*10^-3)

= 2.8753

Answer: 2.88

b)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.12 M

V(NaOH) = 15 mL

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

mol(CH3COOH) = 0.1 M * 25 mL = 2.5 mmol

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

mol(NaOH) = 0.12 M * 15 mL = 1.8 mmol

We have:

mol(CH3COOH) = 2.5 mmol

mol(NaOH) = 1.8 mmol

1.8 mmol of both will react

excess CH3COOH remaining = 0.7 mmol

Volume of Solution = 25 + 15 = 40 mL

[CH3COOH] = 0.7 mmol/40 mL = 0.0175M

[CH3COO-] = 1.8/40 = 0.045M

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 {4.5*10^-2/1.75*10^-2}

= 5.155

Answer: 5.16