a)
find the volume of NaOH used to reach equivalence point
M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)
0.1 M *20.0 mL = 0.2M *V(NaOH)
V(NaOH) = 10 mL
Answer: 10 mL
b)
At half equivalence point, pH = pKa
use:
pKa = -log Ka
= -log (1.8*10^-5)
= 4.7447
So, pH = 4.7447
Answer: 4.74
c)
Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 20 mL
M(NaOH) = 0.2 M
V(NaOH) = 10 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.2 M * 10 mL = 2 mmol
We have:
mol(CH3COOH) = 2 mmol
mol(NaOH) = 2 mmol
2 mmol of both will react to form CH3COO- and H2O
CH3COO- here is strong base
CH3COO- formed = 2 mmol
Volume of Solution = 20 + 10 = 30 mL
Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10
concentration ofCH3COO-,c = 2 mmol/30 mL = 0.0667M
CH3COO- dissociates as
CH3COO- + H2O -----> CH3COOH + OH-
0.0667 0 0
0.0667-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)*6.667*10^-2) = 6.086*10^-6
since c is much greater than x, our assumption is correct
so, x = 6.086*10^-6 M
[OH-] = x = 6.086*10^-6 M
use:
pOH = -log [OH-]
= -log (6.086*10^-6)
= 5.2157
use:
PH = 14 - pOH
= 14 - 5.2157
= 8.7843
Answer: 8.78
d)
Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 20 mL
M(NaOH) = 0.2 M
V(NaOH) = 15 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.2 M * 15 mL = 3 mmol
We have:
mol(CH3COOH) = 2 mmol
mol(NaOH) = 3 mmol
2 mmol of both will react
excess NaOH remaining = 1 mmol
Volume of Solution = 20 + 15 = 35 mL
[OH-] = 1 mmol/35 mL = 0.0286 M
use:
pOH = -log [OH-]
= -log (2.857*10^-2)
= 1.5441
use:
PH = 14 - pOH
= 14 - 1.5441
= 12.4559
Answer: 12.46
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