Question

Determine the pH during the titration of 29.5 mL of 0.324 M acetic acid (K_a = 1.8×10^-5) by 0.414 M NaOH at the following points.

(a) Before the addition of any NaOH ________

(b) After the addition of 5.70 mL of NaOH _______

(c) At the half-equivalence point (the titration midpoint) _______

(d) At the equivalence point _______

(e) After the addition of 34.6 mL of NaOH ________

Answer 1

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

CH3COOH          ----->     H+   + CH3COO-
0.324                 0         0
0.324-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)*0.324) = 2.415*10^-3

since c is much greater than x, our assumption is correct
so, x = 2.415*10^-3 M



use:
pH = -log [H+]
= -log (2.415*10^-3)
= 2.6171
Answer: 2.62

2)when 5.7 mL of NaOH is added
Given:
M(CH3COOH) = 0.324 M
V(CH3COOH) = 29.5 mL
M(NaOH) = 0.414 M
V(NaOH) = 5.7 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.324 M * 29.5 mL = 9.558 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.414 M * 5.7 mL = 2.3598 mmol


We have:
mol(CH3COOH) = 9.558 mmol
mol(NaOH) = 2.3598 mmol

2.3598 mmol of both will react

excess CH3COOH remaining = 7.1982 mmol
Volume of Solution = 29.5 + 5.7 = 35.2 mL
[CH3COOH] = 7.1982 mmol/35.2 mL = 0.2045M

[CH3COO-] = 2.3598/35.2 = 0.067M

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 {6.704*10^-2/0.2045}
= 4.26


Answer: 4.26

3)
At half equivalence point, pH = pKa

use:
pKa = -log Ka
= -log (1.8*10^-5)
= 4.7447

So, pH = 4.7447
Answer: 4.74

4)
find the volume of NaOH used to reach equivalence point
M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)
0.324 M *29.5 mL = 0.414M *V(NaOH)
V(NaOH) = 23.087 mL
Given:
M(CH3COOH) = 0.324 M
V(CH3COOH) = 29.5 mL
M(NaOH) = 0.414 M
V(NaOH) = 23.087 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.324 M * 29.5 mL = 9.558 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.414 M * 23.087 mL = 9.558 mmol


We have:
mol(CH3COOH) = 9.558 mmol
mol(NaOH) = 9.558 mmol

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

CH3COO- here is strong base
CH3COO- formed = 9.558 mmol
Volume of Solution = 29.5 + 23.087 = 52.587 mL
Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10
concentration ofCH3COO-,c = 9.558 mmol/52.587 mL = 0.1818M

CH3COO- dissociates as

CH3COO-        + H2O   ----->     CH3COOH +   OH-
0.1818                        0         0
0.1818-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 ((5.556*10^-10)*0.1818) = 1.005*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.005*10^-5 M



[OH-] = x = 1.005*10^-5 M

use:
pOH = -log [OH-]
= -log (1.005*10^-5)
= 4.9979


use:
PH = 14 - pOH
= 14 - 4.9979
= 9.0021
Answer: 9.00

5)when 34.6 mL of NaOH is added
Given:
M(CH3COOH) = 0.324 M
V(CH3COOH) = 29.5 mL
M(NaOH) = 0.414 M
V(NaOH) = 34.6 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.324 M * 29.5 mL = 9.558 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.414 M * 34.6 mL = 14.3244 mmol


We have:
mol(CH3COOH) = 9.558 mmol
mol(NaOH) = 14.3244 mmol

9.558 mmol of both will react

excess NaOH remaining = 4.7664 mmol
Volume of Solution = 29.5 + 34.6 = 64.1 mL
[OH-] = 4.7664 mmol/64.1 mL = 0.0744 M

use:
pOH = -log [OH-]
= -log (7.436*10^-2)
= 1.1287


use:
PH = 14 - pOH
= 14 - 1.1287
= 12.8713
Answer: 12.87