For 0.50M H2SO3, calculate A) [H+] B) [HSO3-] C) [SO3^2-] D) [H2SO3] K1 for H2SO3 = 1.5 x 10^-2 K2 = 1.0 x 10^-7?

A)
Since K₁ ≫ K₂, then the concentrations of H⁺ and HSO₃⁻ are mainly due to the first dissociation of H₂SO₃, and the effects of the second dissociation on the concentrations of H⁺ and HSO₃⁻ are negligible.

Consider the first dissociation of H₂SO₃.
________ H₂SO₃(aq) ____ ⇌ ____ H⁺(aq) ____ + ____ HSO₃⁻(aq) ______ K₁ = 1.5 * 10⁻²
Initial: ____ 0.50 M _____________ 0 M _____________ 0 M
Change: ___ -x M _____________ +x M ____________ +x M
At eqm: __ (0.50 - x) M __________ x M _____________ x M

At eqm :
K₁ = [H⁺] [HSO₃⁻] / [H₂SO₃]
x² / (0.5 - x) = 1.5 * 10⁻²
x² + (1.5 * 10⁻²)x - (7.5 * 10⁻³) = 0
x = {-(1.5 * 10⁻²) ± √[(1.5 * 10⁻²) + 4(7.5 * 10⁻³)]} / 2
x = 7.9 * 10⁻² M or x = -9.4 * 10⁻² M (rejected)
[H⁺] = x M = 7.9 * 10⁻² M


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B)
Refer to A)
[HSO₃⁻] = x M = 7.9 * 10⁻² M


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C)
Consider the second dissociation of H₂SO₃.
__________ HSO₃⁻(aq) ____ ⇌ ____ H⁺(aq) ____ + ____ SO₃²⁻(aq) ______ K₂ = 1.0 × 10⁻⁷
Initial: ____ 7.9*10⁻² M _______ 7.9*10⁻² M _________ 0 M
Change: ____ -y M ______________ +y M ___________ +y M
At eqm: __ ≈7.9*10⁻² M ______ ≈7.9*10⁻² M _________ y M

K₂ = [H⁺] [SO₃²⁻] / [HSO₃⁻]
(7.9 * 10⁻²) y / (7.9 * 10⁻²) = 1.0 * 10⁻⁷
y = 1.0 * 10⁻⁷ M
[SO₃²⁻] = y M = 1.0 * 10⁻⁷ M

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D)
Refer to A)
[H₂SO₃] = (0.50 - x) M = {0.50 - (7.9 * 10⁻²)} M = 0.42 M

a)
(H^+) = 5*10^-5 M
pH = 4.3
b)
SO3^2- = 5*10^-5 M
c)
(HSO3^-) = Ka2
d)
0.50M H2SO3