Suppose you dissolve 2 g of KH2PO4 in 1L of water and then adjust the pH of the solution to 6.8?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka₁ for H₃PO₄ = 7.1 × 10⁻³
Ka₂ for H₃PO₃ = 6.3 × 10⁻⁸
Ka₃ for H₃PO₃ = 4.2 × 10⁻¹³
Ka values from different sources may be slightly different.

Molar mass of KHPO₄ = 39.1 + 1.0×2 + 31.0 + 16.0×4 = 136.1 g/mol
Initial [HPO₄⁻]ₒ = (2 g) / (136.1 g/mol) = 0.0147 M

pH = -log[H₃O⁺] = 6.8
[H₃O⁺] = 10⁻⁶˙⁸ M and [OH⁻] = 10⁻¹⁴˙⁰/10⁻⁶˙⁸ = 10⁻⁷˙²

Three dissociation reaction occur in the solution:
(1) H₂PO₄⁻(aq) + H₂O(ℓ) ⇌ H₃PO₄(aq) + OH⁻(aq) … Kb = Kw/Ka₁ = 1.41 × 10⁻¹²
(2) H₂PO₄⁻(aq) + H₂O(ℓ) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq) … Ka₂ = 6.3 × 10⁻⁸
(3) HPO₄²⁻(aq) + H₂O(ℓ) ⇌ PO₄³⁻(aq) + H₃O⁺(aq) … Ka₃ = 4.2 × 10⁻¹³

As Ka₂ ≫ Kb > Ka₃, the effects of reactions (1) and (3) are negligible to the molarity determined in reaction (2).

Consider reaction (2) first:
______ H₂PO₄⁻(aq) + H₂O(ℓ) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq) … Ka₂ = 6.3 × 10⁻⁸
Initial: _ 0.0147 M ___________ 0 M _____ 10⁻⁶˙⁸ M
Change: __ -y M ____________ +y M ______ 0 M
Eqm: _ (0.0147 - y) M ________ y M _____ 10⁻⁶˙⁸ M

At equilibrium:
Ka₂ = [HPO₄²⁻] [H₃O⁺] / [H₂PO₄⁻]
6.3 × 10⁻⁸ = y × 10⁻⁶˙⁸ / 0.0147 - y
9.261 × 10⁻¹⁰ - (6.3 × 10⁻⁸)y = (10⁻⁶˙⁸)y
[(10⁻⁶˙⁸) + (6.3 × 10⁻⁸)] y = 9.261 × 10⁻¹⁰
y = 9.261 × 10⁻¹⁰ / [(10⁻⁶˙⁸) + (6.3 × 10⁻⁸)] = 4.18 × 10⁻³
[HPO₄²⁻] = 4.18 × 10⁻³ M
[H₂PO₄⁻] = [0.0147 - (4.18 × 10⁻³)] = 1.05 × 10⁻² M

Consider reaction (1):
________ H₂PO₄⁻(aq) + H₂O(ℓ) ⇌ H₃PO₄(aq) + OH⁻(aq) … Kb = 1.41 × 10⁻¹²
Eqm: _ ~1.05 × 10⁻² M _________ ? M ______ 10⁻⁷˙²

At equilibrium:
Kb = [H₃PO₄] [OH⁻] / [H₂PO₄⁻]
1.41 × 10⁻¹² = [H₃PO₄] × 10⁻⁷˙² / (1.05 × 10⁻²)
[H₃PO₄] = (1.41 × 10⁻¹²) × (1.05 × 10⁻²) / 10⁻⁷˙² = 2.35 × 10⁻⁷ M

Consider reaction (3):
______ HPO₄²⁻(aq) + H₂O(ℓ) ⇌ PO₄³⁻(aq) + H₃O⁺(aq) … Ka₃ = 4.2 × 10⁻¹³
Eqm: ~4.18 × 10⁻³ M _________ ? M ____ 10⁻⁶˙⁸ M

At equilibrium:
Ka₃ = [PO₄³⁻] [H₃O⁺] / [HPO₄²⁻]
4.2 × 10⁻¹³ = [PO₄³⁻] ×10⁻⁶˙⁸ / (4.18 × 10⁻³)
[PO₄³⁻] = (4.2 × 10⁻¹³) × (4.18 × 10⁻³) / 10⁻⁶˙⁸ = 1.11 × 10⁻⁸ M

The answers:
[H₃PO₄] = 2.35 × 10⁻⁷ M
[H₂PO₄⁻] = 1.05 × 10⁻² M
[HPO₄²⁻] = 4.18 × 10⁻³ M
[PO₄³⁻] = 1.11 × 10⁻⁸ M