A) 2.7218
B) 4.2676
C) 4.7447
D) 5.6990
E) 8.7843
F) 11.5086
Explanation:
A)
When 0.0 mL of KOH is added
,
CH3COOH dissociates as:
CH3COOH ⇒ H⁺ + CH3COO⁻
0.2 0 0
0.2-x x x
Kₐ = [H⁺][CH3COO⁻] / [CH3COOH]
Kₐ = x*x / (c-x)
Assuming x can be ignored as compared to c, the above expression becomes
Kₐ = x*x / c
so, x = √Kₐ.c
x = √((1.8*10⁻⁵)*0.2) = 1.897*10⁻³
since c is much greater than x, then our assumption is correct
so, x = 1.897*10⁻³ M
use:
pH = -log [H⁺]
= -log (1.897*10⁻³)
= 2.7218
B)
When 50.0 mL of KOH is added
Given:
M(CH3COOH) = 0.2 M
V(CH3COOH) = 100 mL
M(KOH) = 0.1 M
V(KOH) = 50 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.2 M * 100 mL
= 20 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 50 mL = 5 mmol
We have:
mol(CH3COOH) = 20 mmol
mol(KOH) = 5 mmol
5 mmol of both will react
excess CH3COOH remaining = 15 mmol
Volume of Solution = 100 + 50 = 150 mL
[CH3COOH] = 15 mmol / 150 mL = 0.1M
[CH3COO⁻] = 5 / 150 = 0.0333M
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 {3.333*10⁻² / 0.1}
= 4.268
C)
When 100.0 mL of KOH is added
Given:
M(CH3COOH) = 0.2 M
V(CH3COOH) = 100 mL
M(KOH) = 0.1 M
V(KOH) = 100 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.2 M * 100 mL = 20 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 100 mL = 10 mmol
We have:
mol(CH3COOH) = 20 mmol
mol(KOH) = 10 mmol
10 mmol of both will react
excess CH3COOH remaining = 10 mmol
Volume of Solution = 100 + 100 = 200 mL
[CH3COOH] = 10 mmol / 200 mL = 0.05M
[CH3COO-] = 10/200 = 0.05M
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 (5*10⁻² / 5*10⁻²)
= 4.745
D)
When 180.0 mL of KOH is added
Given:
M(CH3COOH) = 0.2 M
V(CH3COOH) = 100 mL
M(KOH) = 0.1 M
V(KOH) = 180 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.2 M * 100 mL = 20 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 180 mL = 18 mmol
We have:
mol(CH3COOH) = 20 mmol
mol(KOH) = 18 mmol
18 mmol of both will react
excess CH3COOH remaining = 2 mmol
Volume of Solution = 100 + 180 = 280 mL
[CH3COOH] = 2 mmol / 280 mL = 0.0071M
[CH3COO-] = 18/280 = 0.0643M
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 (6.429*10⁻² / 7.143*10⁻³)
= 5.699
E)
When 200.0 mL of KOH is added
Given:
M(CH3COOH) = 0.2 M
V(CH3COOH) = 100 mL
M(KOH) = 0.1 M
V(KOH) = 200 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.2 M * 100 mL = 20 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 200 mL = 20 mmol
We have:
mol(CH3COOH) = 20 mmol
mol(KOH) = 20 mmol
20 mmol of both will react to form CH3COO⁻ and H₂O
CH3COO⁻ here is strong base
CH3COO⁻ formed = 20 mmol
Volume of Solution = 100 + 200 = 300 mL
Kb of CH3COO- = Kw / Ka = 1*10⁻¹⁴ / 1.8*10⁻⁵ = 5.556*10⁻¹⁰
concentration of CH3COO⁻, c = 20 mmol / 300 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 = √Kb.c
x = √((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
F)
When 210.0 mL of KOH is added
Given:
M(CH3COOH) = 0.2 M
V(CH3COOH) = 100 mL
M(KOH) = 0.1 M
V(KOH) = 210 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.2 M * 100 mL = 20 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 210 mL = 21 mmol
We have:
mol(CH3COOH) = 20 mmol
mol(KOH) = 21 mmol
20 mmol of both will react
excess KOH remaining = 1 mmol
Volume of Solution = 100 + 210 = 310 mL
[OH⁻] = 1 mmol / 310 mL = 0.0032 M
use:
pOH = -log [OH⁻]
= -log (3.226*10⁻³)
= 2.4914
use:
PH = 14 - pOH
= 14 - 2.4914
= 11.5086
