Question

Given the equilibrium reactions:

#1) SO₂(aq) + H₂O(l) = HSO₃^- (aq) + H⁺(aq)

#2) HSO₃^- (aq) + H⁺(aq) = SO₃^2- + 2H⁺(aq)

What is the equilibrium constant for reaction #1 if the concentration of SO₂ is 8.50E-20 M, SO₃^2- is 0.619 M, and H⁺ is 2.00E-07 M? You know that the equilibrium constant (K) for reaction #2 is 130.

Answer 1

Reaction 1) SO2(aq) + H2O(l) HSO3- (aq) + H+(aq)

Reaction 2) HSO3- (aq) + H+(aq) SO32- + 2H+(aq)

for reaction 2

Kc = [SO3 2-][H+]²/[HSO 3-][H+]

130 = 0.619 x(2.0 x 10^-7)²/[HSO 3-](2.0 x 10^-7)

  [HSO 3-] = 0.619 x(2.0 x 10^-7)/130 = 1.238 x 10^-7/130 = 0.00952 x 10^-7M

For reaction 1

Kc =  [HSO 3-][H+]/[SO2][H2O]

Kc =  [HSO 3-][H+]/[SO2] (here H2O is solvent so its concentration is taken as unity)

Kc = 0.00952 x10^-7x (2.0 x 10^-7)/(8.50 x 10^-20)

= 0.01904x 10^-14 /8.50 x 10^-20 = 0.00224 x 10⁶