What is the pH of a solution containing 1.396 mol L-1 of a weak acid with pKA = 8.39 and 0.506 mol L-1 of another weak acid with pKA = 4.70?

Method 1 : Use approximation

pKa = -log(Ka) and thus Ka = 10^(-pKa)
Ka(first acid) = 10⁻⁸˙³⁹ = 4.07 × 10⁻⁹
Ka(second acid) = 10⁻⁴˙⁷⁰ = 2.00 × 10⁻⁵

As Ka(second acid) ≫ Ka(first acid), H₃O⁺ in the solution is almost entirely formed from the dissociation of the second acid.
Hence, we can neglect the dissociation of the first acid.

Denote the second acid as HA.
_______________ HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq) …… Ka = 2.00 × 10⁻⁵
Initial: ________ 0.506 M _________ 0 M ____ 0 M
Change: ________ -a M ___________ +a M ___ +a M
Equilibrium: __ (0.506 - a) M _______ a M ____ a M

As Ka is very small, the dissociation of HA is to a very small extent.
We can assume that 0.506 ≫ a
Hence, [HB] at equilibrium = (0.506 - a) M ≈ 0.506 M

At equilibrium: Ka = [A⁻] [H₃O⁺] / [HA]
2.00 × 10⁻⁵ = a² / 0.506
a = √{(2.00 × 10⁻⁵) × 0.506} = 0.00318

pH = -log[H₃O⁺] = -log(a) = -log(0.00318) = 2.50

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Method 2 : Consider the dissociations of both the acids.

pKa = -log(Ka) and thus Ka = 10^(-pKa)
Ka(first acid) = 10⁻⁸˙³⁹ = 4.07 × 10⁻⁹
Ka(second acid) = 10⁻⁴˙⁷⁰ = 2.00 × 10⁻⁵

Denote the first acid as HX.
_______________ HX(aq) + H₂O(l) ⇌ X⁻(aq) + H₃O⁺(aq) …… Ka(HX) = 4.07 × 10⁻⁹
Initial: ________ 1.396 M _________ 0 M ____ 0 M
Change: ________ -x M __________ +x M __ +(x + y) M
Equilibrium: __ (1.396 - x) M ______ x M ____ (x + y) M

As Ka(HX) is very small, the dissociation of HX is to a very small extent.
We can assume that 1.396 ≫ x
Hence, [HX] at equilibrium = (1.396 - y) M ≈ 1.396 M

At equilibrium: Ka(HX) = [X⁻] [H₃O⁺] / [HX]
4.07 × 10⁻⁹ = x (x + y) / 1.396
x (x + y) = (4.07 × 10⁻⁹) × 1.396
x² + xy = 5.68 × 10⁻⁹ …… {1}

Denote the second acid as HY.
_______________ HY(aq) + H₂O(l) ⇌ Y⁻(aq) + H₃O⁺(aq) …… Ka(HY) = 2.00 × 10⁻⁵
Initial: ________ 0.506 M _________ 0 M ____ 0 M
Change: ________ -y M __________ +y M __ +(x + y) M
Equilibrium: __ (0.506 - y) M _______ y M ___ (x + y) M

As Ka(HY) is very small, the dissociation of HY is to a very small extent.
We can assume that 0.506 ≫ y
Hence, [HY] at equilibrium = (0.506 - y) M ≈ 0.506 M

At equilibrium: Ka(HY) = [Y⁻] [H₃O⁺] / [HY]
2.00 × 10⁻⁵ = y (x + y) / 0.506
y (x + y) = (2.00 × 10⁻⁵) × 0.506
xy + y² = 1.01 × 10⁻⁵ …… {2}

{1} + {2}:
x² + 2xy + y² = 1.01 × 10⁻⁵
(x + y)² = 1.01 × 10⁻⁵
x + y = √(1.01 × 10⁻⁵) = 0.00318
pH = -log[H₃O⁺] = -log(x+y) = -log(0.00318) = 2.50