Question

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

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

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

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