Chemistry, help? Acid/Base solutions. Thanks!?

For Kₐ₁ ≫ Kₐ₂ ≫ Kₐ₃
pH almost completely depends on the first dissociation only.


The first ionization :
H₃PO₄(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + H₃O⁺(aq) ...... Kₐ₁ = 7.5 × 10⁻³

At equilibrium :
[H₃PO₄] = (7.75 - x) M
[H₂PO₄⁻] = [H₃O⁺] = x M

Kₐ₁ = [H₂PO₄⁻] [H₃O⁺] / [H₃PO₄]
7.5 × 10⁻³ = x² / (7.75 - x)
x² + (7.5 × 10⁻³)x - (7.75 × 7.5 × 10⁻³) = 0
Solve the equation. x = 0.237

[H₃PO₄] = (7.75 - 0.237) M = 7.51 M
[H₂PO₄⁻] = 0.237 M
[H₃O⁺] = 0.237 M

pH = -log[H₃O⁺] = -log(0.237) = 0.63


The second dissociation :
H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq) ...... Kₐ₂ = 6.2 × 10⁻⁸

At equilibrium :
[H₂PO₄⁻] ≈ 0.237 M and [H₃O⁺] ≈ 0.237 M (for Kₐ₁ ≫ Kₐ₂)

Kₐ₂ = [HPO₄²⁻] [H₃O⁺] / [H₂PO₄⁻]
6.2 × 10⁻⁸ = [HPO₄²⁻] × 0.237 / 0.237
[HPO₄²⁻] = 6.2 × 10⁻⁸ M


The third dissociation :
HPO₄²⁻(aq) + H₂O(l) ⇌ PO₄³⁻(aq) + H₃O⁺(aq) ...... Kₐ₃ = 4.8 × 10⁻¹³

At equilibrium :
[HPO₄²⁻] ≈ 6.2 × 10⁻⁸ M (for Kₐ₂ ≫ Kₐ₃)
[H₃O⁺] ≈ 0.237 M (for Kₐ₁ ≫ Kₐ₃)

Kₐ₃ = [PO₄³⁻] [H₃O⁺] / [HPO₄²⁻]
4.8 × 10⁻¹³ = [PO₄³⁻] × 0.237 / (6.2 × 10⁻⁸)
[PO₄³⁻] = 1.26 × 10⁻¹⁹ M

Ka1 = [H+][H2PO4-] / [ H3PO4]

NB [H+] = [ H2PO3-]
Hence
Ka1 = [H+]^2 /[H3PO4]
[H+]^2 = Ka1 x [H3PO4]
[H+]^2 = 7.5 x 10^-3 x 7.75
[H+]^2 = 0.058125
[H+] = sqrt(0.058125) = 0.24109
pH = -log(10)[0.24109)
pH = --0.6178
pH = 0.62 ( 2d.p)

Ka2 = [H+]^2[HPO4^2-] / [ H3PO4]

NB [H+] + [H+] = [ HPO3^2-]

Hence
Ka2 = [H+]^3 /[H3PO4]
[H+]^3 = Ka2x [H3PO4]
[H+]^3 = 6.2 x 10^-8 x 7.75
[H+]^3 =4.805 x 10^-7
[H+] = cu.rt(4.805 x 10^-7) = 7.8032 x 10^-3

pH = -log(10)[7.8032 x 10^-3]
pH = --2.106...
pH = 2.11 ( 2d.p)

Ka1 = [H+][H2PO4-] / [ H3PO4]

NB [H+]^3 = [ PO3^3-]
Hence
Ka3 = [H+]^4 /[H3PO4]
[H+]^4 = Ka3 x [H3PO4]
[H+]^4 = 4.8 x 10^-13 X 7.75

[H+]^4 =3.72 x 10^-12
[H+] = 4th.rt(3.72 x 10^-12) = 1.38878 x 10^-3
pH = -log(10)[1.38878 x 10^-3]
pH = --- 2.85736...
pH = 2.86 ( 2d.p)