Question

Calculate the pH during titration of 75.0mL of 0.100M benzoic acid by 0.100M NaOH before the addition of NaOH and after 20mL NaOH are added. Ka=6.3x10^-5

Answer 1

1)when 0.0 mL of NaOH is added

C6H5COOH dissociates as:

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

0.1 0 0

0.1-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 ((6.3*10^-5)*0.1) = 2.51*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)

6.3*10^-5 = x^2/(0.1-x)

6.3*10^-6 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-6.3*10^-6 = 0

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

a = 1

b = 6.3*10^-5

c = -6.3*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 = 2.52*10^-5

roots are :

x = 2.479*10^-3 and x = -2.542*10^-3

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

x = 2.479*10^-3

use:

pH = -log [H+]

= -log (2.479*10^-3)

= 2.6058

Answer: 2.61

2)when 20.0 mL of NaOH is added

Given:

M(C6H5COOH) = 0.1 M

V(C6H5COOH) = 75 mL

M(NaOH) = 0.1 M

V(NaOH) = 20 mL

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

mol(C6H5COOH) = 0.1 M * 75 mL = 7.5 mmol

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

mol(NaOH) = 0.1 M * 20 mL = 2 mmol

We have:

mol(C6H5COOH) = 7.5 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

excess C6H5COOH remaining = 5.5 mmol

Volume of Solution = 75 + 20 = 95 mL

[C6H5COOH] = 5.5 mmol/95 mL = 0.0579M

[C6H5COO-] = 2/95 = 0.0211M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.3*10^-5

pKa = - log (Ka)

= - log(6.3*10^-5)

= 4.201

use:

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

= 4.201+ log {2.105*10^-2/5.789*10^-2}

= 3.761

Answer: 3.76