Question

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 mL sample of 0.109 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.109 M NaOH solution. (Assume that all solutions are at 25°C.) (a) Calculate the pH after the addition of the following amounts of NaOH. Vol. NaOH added pH 0 mL 4.0 mL 8.0 mL 12.5 mL 20.0 mL 24.0 mL 24.5 mL 24.9 mL 25.0 mL 25.1 mL 26.0 mL 28.0 mL 30.0 mL

Answer 1

Given:
pKa = 3.86

use:
pKa = -log Ka
3.86 = -log Ka
Ka = 1.38*10^-4

1)when 0.0 mL of NaOH is added
HC3H5O3 dissociates as:

HC3H5O3          ----->     H+   + C3H5O3-
0.109                 0         0
0.109-x               x         x


Ka = [H+][C3H5O3-]/[HC3H5O3]
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 ((1.38*10^-4)*0.109) = 3.879*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)
1.38*10^-4 = x^2/(0.109-x)
1.505*10^-5 - 1.38*10^-4 *x = x^2
x^2 + 1.38*10^-4 *x-1.505*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.38*10^-4
c = -1.505*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 = 6.02*10^-5

roots are :
x = 3.811*10^-3 and x = -3.949*10^-3

since x can't be negative, the possible value of x is
x = 3.811*10^-3

use:
pH = -log [H+]
= -log (3.811*10^-3)
= 2.419
Answer: 2.42

2)when 4.0 mL of NaOH is added
Given:
M(HC3H5O3) = 0.109 M
V(HC3H5O3) = 25 mL
M(NaOH) = 0.109 M
V(NaOH) = 4 mL


mol(HC3H5O3) = M(HC3H5O3) * V(HC3H5O3)
mol(HC3H5O3) = 0.109 M * 25 mL = 2.725 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.109 M * 4 mL = 0.436 mmol


We have:
mol(HC3H5O3) = 2.725 mmol
mol(NaOH) = 0.436 mmol

0.436 mmol of both will react

excess HC3H5O3 remaining = 2.289 mmol
Volume of Solution = 25 + 4 = 29 mL
[HC3H5O3] = 2.289 mmol/29 mL = 0.0789M

[C3H5O3-] = 0.436/29 = 0.015M

They form acidic buffer
acid is HC3H5O3
conjugate base is C3H5O3-


Ka = 1.38*10^-4

pKa = - log (Ka)
= - log(1.38*10^-4)
= 3.86

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.86+ log {1.503*10^-2/7.893*10^-2}
= 3.14


Answer: 3.14

3)when 8.0 mL of NaOH is added
Given:
M(HC3H5O3) = 0.109 M
V(HC3H5O3) = 25 mL
M(NaOH) = 0.109 M
V(NaOH) = 8 mL


mol(HC3H5O3) = M(HC3H5O3) * V(HC3H5O3)
mol(HC3H5O3) = 0.109 M * 25 mL = 2.725 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.109 M * 8 mL = 0.872 mmol


We have:
mol(HC3H5O3) = 2.725 mmol
mol(NaOH) = 0.872 mmol

0.872 mmol of both will react

excess HC3H5O3 remaining = 1.853 mmol
Volume of Solution = 25 + 8 = 33 mL
[HC3H5O3] = 1.853 mmol/33 mL = 0.0562M

[C3H5O3-] = 0.872/33 = 0.0264M

They form acidic buffer
acid is HC3H5O3
conjugate base is C3H5O3-


Ka = 1.38*10^-4

pKa = - log (Ka)
= - log(1.38*10^-4)
= 3.86

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.86+ log {2.642*10^-2/5.615*10^-2}
= 3.533


Answer: 3.53

4)when 12.5 mL of NaOH is added
Given:
M(HC3H5O3) = 0.109 M
V(HC3H5O3) = 25 mL
M(NaOH) = 0.109 M
V(NaOH) = 12.5 mL


mol(HC3H5O3) = M(HC3H5O3) * V(HC3H5O3)
mol(HC3H5O3) = 0.109 M * 25 mL = 2.725 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.109 M * 12.5 mL = 1.3625 mmol


We have:
mol(HC3H5O3) = 2.725 mmol
mol(NaOH) = 1.3625 mmol

1.3625 mmol of both will react

excess HC3H5O3 remaining = 1.3625 mmol
Volume of Solution = 25 + 12.5 = 37.5 mL
[HC3H5O3] = 1.3625 mmol/37.5 mL = 0.0363M

[C3H5O3-] = 1.3625/37.5 = 0.0363M

They form acidic buffer
acid is HC3H5O3
conjugate base is C3H5O3-


Ka = 1.38*10^-4

pKa = - log (Ka)
= - log(1.38*10^-4)
= 3.86

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.86+ log {3.633*10^-2/3.633*10^-2}
= 3.86


Answer: 3.86

Only 4 parts at a time please