Question

9. Acetic acid (CH3CO2H) has a pk of 4.76 at 25°C. A 5.00 mL aliquot of a 0.0185 M solution of acetic acid is titrated with a 0.1000 M standardized solution of NaOH. (a) What is the pH of the acetic acid solution before any of the standardized solution is added? (b) What volume of the standardized solution is necessary to reach the equiv- alence point? (c) What is the pH of the solution after 1.48 mL of the standardized solution is added?

Answer 1

Given:
pKa = 4.76

use:
pKa = -log Ka
4.76 = -log Ka
Ka = 1.738*10^-5

a)when 0.0 mL of NaOH is added
CH3COOH dissociates as:

CH3COOH          ----->     H+   + CH3COO-
1.85*10^-2                 0         0
1.85*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 ((1.738*10^-5)*1.85*10^-2) = 5.67*10^-4

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.738*10^-5 = x^2/(1.85*10^-2-x)
3.215*10^-7 - 1.738*10^-5 *x = x^2
x^2 + 1.738*10^-5 *x-3.215*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.738*10^-5
c = -3.215*10^-7

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 = 1.286*10^-6

roots are :
x = 5.584*10^-4 and x = -5.758*10^-4

since x can't be negative, the possible value of x is
x = 5.584*10^-4

use:
pH = -log [H+]
= -log (5.584*10^-4)
= 3.2531
Answer: 3.25

b)
find the volume of NaOH used to reach equivalence point
M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)
0.0185 M *5.0 mL = 0.1M *V(NaOH)
V(NaOH) = 0.925 mL
Given:
M(CH3COOH) = 0.0185 M
V(CH3COOH) = 5 mL
M(NaOH) = 0.1 M
V(NaOH) = 0.925 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.0185 M * 5 mL = 0.0925 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 0.925 mL = 0.0925 mmol


We have:
mol(CH3COOH) = 0.0925 mmol
mol(NaOH) = 0.0925 mmol

0.0925 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base
CH3COO- formed = 0.0925 mmol
Volume of Solution = 5 + 0.925 = 5.925 mL
Kb of CH3COO- = Kw/Ka = 1*10^-14/1.738*10^-5 = 5.754*10^-10
concentration ofCH3COO-,c = 0.0925 mmol/5.925 mL = 0.0156M

CH3COO- dissociates as

CH3COO-        + H2O   ----->     CH3COOH +   OH-
0.0156                        0         0
0.0156-x                      x         x


Kb = [CH3COOH][OH-]/[CH3COO-]
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 ((5.754*10^-10)*1.561*10^-2) = 2.997*10^-6

since c is much greater than x, our assumption is correct
so, x = 2.997*10^-6 M



[OH-] = x = 2.997*10^-6 M

use:
pOH = -log [OH-]
= -log (2.997*10^-6)
= 5.5233


use:
PH = 14 - pOH
= 14 - 5.5233
= 8.4767
Answer: 8.48

c)when 1.48 mL of NaOH is added
Given:
M(CH3COOH) = 0.0185 M
V(CH3COOH) = 5 mL
M(NaOH) = 0.1 M
V(NaOH) = 1.48 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.0185 M * 5 mL = 0.0925 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 1.48 mL = 0.148 mmol


We have:
mol(CH3COOH) = 0.0925 mmol
mol(NaOH) = 0.148 mmol

0.0925 mmol of both will react

excess NaOH remaining = 0.0555 mmol
Volume of Solution = 5 + 1.48 = 6.48 mL
[OH-] = 0.0555 mmol/6.48 mL = 0.0086 M

use:
pOH = -log [OH-]
= -log (8.565*10^-3)
= 2.0673


use:
PH = 14 - pOH
= 14 - 2.0673
= 11.9327
Answer: 11.93