Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
![K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}](https://tex.z-dn.net/?f=%20K_%7Ba1%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BHSO_%7B3%7D%5E%7B-%7D%5D%7D%7B%5BH_%7B2%7DSO_%7B3%7D%5D%7D%20%3D%20%5Cfrac%7Bx%5Ccdot%20x%7D%7B0.5%20-%20x%7D%20%3D%20%5Cfrac%20%7Bx%5E%7B2%7D%7D%7B0.5%20-x%7D%20)
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
![[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M](https://tex.z-dn.net/?f=%20%5BHSO_%7B3%7D%5E%7B-%7D%5D%20%3D%20%5BH%5E%7B%2B%7D%5D%20%3D%207.94%20%5Ccdot%2010%5E%7B-2%7DM%20)
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
![K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}](https://tex.z-dn.net/?f=%20K_%7Ba2%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BSO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHSO_%7B3%7D%5E%7B-%7D%5D%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B7.94%20%5Ccdot%2010%5E%7B-2%7D%20-%20x%7D%20)
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
![[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M](https://tex.z-dn.net/?f=%20%5BSO_%7B3%7D%5E%7B2-%7D%5D%20%3D%20%5BH%5E%7B%2B%7D%5D%20%3D%207.07%20%5Ccdot%2010%5E%7B-5%7DM%20)
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
I hope it helps you!
