When a 23.8 mL sample of a
0.491 M aqueous acetic acid
solution is titrated with a 0.381 M aqueous
potassium hydroxide solution,
(1) What is the pH at the midpoint in the titration?
(2) What is the pH at the equivalence point of the titration?
(3) What is the pH after 46.0 mL of
potassium hydroxide have been added
1)
At half equivalence point, pH = pKa
use:
pKa = -log Ka
= -log (1.8*10^-5)
= 4.7447
So, pH = 4.7447
Answer: 4.74
2)
find the volume of KOH used to reach equivalence point
M(CH3COOH)*V(CH3COOH) =M(KOH)*V(KOH)
0.491 M *23.8 mL = 0.381M *V(KOH)
V(KOH) = 30.6714 mL
Given:
M(CH3COOH) = 0.491 M
V(CH3COOH) = 23.8 mL
M(KOH) = 0.381 M
V(KOH) = 30.6714 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.491 M * 23.8 mL = 11.6858 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.381 M * 30.6714 mL = 11.6858 mmol
We have:
mol(CH3COOH) = 11.6858 mmol
mol(KOH) = 11.6858 mmol
11.6858 mmol of both will react to form CH3COO- and H2O
CH3COO- here is strong base
CH3COO- formed = 11.6858 mmol
Volume of Solution = 23.8 + 30.6714 = 54.4714 mL
Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10
concentration ofCH3COO-,c = 11.6858 mmol/54.4714 mL = 0.2145M
CH3COO- dissociates as
CH3COO- + H2O
-----> CH3COOH + OH-
0.2145
0 0
0.2145-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)*0.2145) = 1.092*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.092*10^-5 M
[OH-] = x = 1.092*10^-5 M
use:
pOH = -log [OH-]
= -log (1.092*10^-5)
= 4.9619
use:
PH = 14 - pOH
= 14 - 4.9619
= 9.0381
Answer: 9.04
3)
Given:
M(CH3COOH) = 0.491 M
V(CH3COOH) = 23.8 mL
M(KOH) = 0.381 M
V(KOH) = 46 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.491 M * 23.8 mL = 11.6858 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.381 M * 46 mL = 17.526 mmol
We have:
mol(CH3COOH) = 11.6858 mmol
mol(KOH) = 17.526 mmol
11.6858 mmol of both will react
excess KOH remaining = 5.8402 mmol
Volume of Solution = 23.8 + 46 = 69.8 mL
[OH-] = 5.8402 mmol/69.8 mL = 0.0837 M
use:
pOH = -log [OH-]
= -log (8.367*10^-2)
= 1.0774
use:
PH = 14 - pOH
= 14 - 1.0774
= 12.9226
Answer: 12.92
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