Question

15. (25 pts) If 50 mL of 0.20 M benzoic acid (K, = 6.5 x 10) is titrated with 0.25 M NaOH, what is the pH after 0 ml, 10 ml, 20 ml, 40 ml and 50 ml of base have been added? (You may use the approximation.). BE SURE TO SHOW ALL CALCULATIONS.

Answer 1

1)when 0.0 mL of NaOH is added

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.2 0 0

0.2-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

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 ((6.5*10^-5)*0.2) = 3.606*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)

6.5*10^-5 = x^2/(0.2-x)

1.3*10^-5 - 6.5*10^-5 *x = x^2

x^2 + 6.5*10^-5 *x-1.3*10^-5 = 0

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

a = 1

b = 6.5*10^-5

c = -1.3*10^-5

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 = 5.2*10^-5

roots are :

x = 3.573*10^-3 and x = -3.638*10^-3

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

x = 3.573*10^-3

use:

pH = -log [H+]

= -log (3.573*10^-3)

= 2.4469

Answer: 2.45

2)when 10.0 mL of NaOH is added

Given:

M(C6H5COOH) = 0.2 M

V(C6H5COOH) = 50 mL

M(NaOH) = 0.25 M

V(NaOH) = 10 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.2 M * 50 mL = 10 mmol

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

mol(NaOH) = 0.25 M * 10 mL = 2.5 mmol

We have:

mol(C6H5COOH) = 10 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess C6H5COOH remaining = 7.5 mmol

Volume of Solution = 50 + 10 = 60 mL

[C6H5COOH] = 7.5 mmol/60 mL = 0.125M

[C6H5COO-] = 2.5/60 = 0.0417M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.5*10^-5

pKa = - log (Ka)

= - log(6.5*10^-5)

= 4.187

use:

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

= 4.187+ log {4.167*10^-2/0.125}

= 3.71

Answer: 3.71

3)when 20.0 mL of NaOH is added

Given:

M(C6H5COOH) = 0.2 M

V(C6H5COOH) = 50 mL

M(NaOH) = 0.25 M

V(NaOH) = 20 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.2 M * 50 mL = 10 mmol

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

mol(NaOH) = 0.25 M * 20 mL = 5 mmol

We have:

mol(C6H5COOH) = 10 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react

excess C6H5COOH remaining = 5 mmol

Volume of Solution = 50 + 20 = 70 mL

[C6H5COOH] = 5 mmol/70 mL = 0.0714M

[C6H5COO-] = 5/70 = 0.0714M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.5*10^-5

pKa = - log (Ka)

= - log(6.5*10^-5)

= 4.187

use:

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

= 4.187+ log {7.143*10^-2/7.143*10^-2}

= 4.187

Answer: 4.19

4)when 40.0 mL of NaOH is added

Given:

M(C6H5COOH) = 0.2 M

V(C6H5COOH) = 50 mL

M(NaOH) = 0.25 M

V(NaOH) = 40 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.2 M * 50 mL = 10 mmol

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

mol(NaOH) = 0.25 M * 40 mL = 10 mmol

We have:

mol(C6H5COOH) = 10 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base

C6H5COO- formed = 10 mmol

Volume of Solution = 50 + 40 = 90 mL

Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.5*10^-5 = 1.538*10^-10

concentration ofC6H5COO-,c = 10 mmol/90 mL = 0.1111M

C6H5COO- dissociates as

C6H5COO- + H2O -----> C6H5COOH + OH-

0.1111 0 0

0.1111-x x x

Kb = [C6H5COOH][OH-]/[C6H5COO-]

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 ((1.538*10^-10)*0.1111) = 4.134*10^-6

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

so, x = 4.134*10^-6 M

[OH-] = x = 4.134*10^-6 M

use:

pOH = -log [OH-]

= -log (4.134*10^-6)

= 5.3836

use:

PH = 14 - pOH

= 14 - 5.3836

= 8.6164

Answer: 8.62

5)when 50.0 mL of NaOH is added

Given:

M(C6H5COOH) = 0.2 M

V(C6H5COOH) = 50 mL

M(NaOH) = 0.25 M

V(NaOH) = 50 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.2 M * 50 mL = 10 mmol

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

mol(NaOH) = 0.25 M * 50 mL = 12.5 mmol

We have:

mol(C6H5COOH) = 10 mmol

mol(NaOH) = 12.5 mmol

10 mmol of both will react

excess NaOH remaining = 2.5 mmol

Volume of Solution = 50 + 50 = 100 mL

[OH-] = 2.5 mmol/100 mL = 0.025 M

use:

pOH = -log [OH-]

= -log (2.5*10^-2)

= 1.6021

use:

PH = 14 - pOH

= 14 - 1.6021

= 12.3979

Answer: 12.40