Answer: (1). pH = 1.70
(2). pH = 2.3
(3). pH = 3.3
(4). pH = 4.3
(5). pH = 8.41
(6). pH = 10.22
Explanation:
we assume that the formula representation of acid is H₂A
the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;
H₂A + OH⁻ → HA⁻ + H₂O
HA⁻ + OH⁻ → A²⁻ + H₂O
the volume of NaOH (V₀) at the first equivalent point is,
V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL
where volume of NaOH at 1/2 equivalent point is,
V₀/2 = 10.0mL
also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL
the volume of NaOH at 1/2 second equivalent point is,
V₀ + V₀/2 = 30.0mL
Volume of NaOH after second equivalence exceeds 40mL
therefore, at 0 mL NaOH addition;
where the extent of ionization is assumed to be x, we have
H₂A ⇆ HA⁻ + H⁺
where initial: 0.1 M - -
change: -x +x +x
Equili: 0.1-x x x
Kаl = [HA⁻][H⁺] / [H₂A]
10⁻²³ = (x)(x) / (0.1-x)
x = 0.020
[H⁺] = 0.020 M
pH = -log [H⁺]
pH = -log(0.020)
pH = 1.70
(2). at 10 mL NaOH addition
[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol
[OH⁻] = 0.1 M * 10 mL = 1 mmol
after reaction:
[H₂A] = 1 mmol
[H⁻] = 1 mmol
pH = pKa₁ + log [HA⁻] / [[HA⁻]
pH = 2.3 + log 1mmol / 1mmol
pH = 2.3
(3). pH at the first equivalence point is,
pH = 1/2 (pKa₁ + pKa₂)
pH = 1/2(2.3 + 4.3) = 3.3
pH = 3.3
(4). pH at the second 1/2 equivalence point is
pH = pKa₂ = 4.3
pH = 4.3
(5). pH at the second equivalence point;
all H₂A is converted into A²⁻
[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M
at equilibrium:
A²⁻ + H²O ⇆ HA⁺ OH⁻
0.033 - x
from the Kb₁ expression,
Kb₁ = [OH⁻][HA⁻] / [A²]
Kw/Ka₂ = x²/(0.0333 - x)
10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)
x = 2.57 * 10⁻⁶
[OH⁻] = 2.57 * 10⁻⁶M
pH = -log Kw/[OH⁻] = 8.41
pH = 8.41
(6). pH after second equivalence point;
assuming the volume of NaOH is 40.10 mL
after second equivalence point OH⁻ in excess
[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M
pH = 0=-log Kw/[OH⁻] = 10.22
pH = 10.22
