Respuesta :
Explanation:
As we know that [tex]H_{2}SO_{3}[/tex] is a strong acid as compared to carbonic acid. As a result, sulfurous acid ([tex]H_{2}SO_{3}[/tex]) is a stronger acid and it will over come carbonic acid.
In a neutralization reaction, the equilibrium will favor side of weaker acids and bases in neutralization.
As [tex]K_{a}[/tex] of [tex]H_{2}SO_{3}[/tex] is greater than the [tex]K_{a}[/tex] value of [tex]H_{2}CO_{3}[/tex].
Thus, we can conclude that the reaction equilibrium will move towards the right hand side.
The equilibrium of the reaction H₂SO₃(aq) + HCO₃⁻(aq) ⇄ HSO₃⁻(aq) + H₂CO₃(aq) will lie to the right side, to the formation of products.
The reaction is the following:
H₂SO₃(aq) + HCO₃⁻(aq) ⇄ HSO₃⁻(aq) + H₂CO₃(aq) (1)
The first acid constant (H₂SO₃) is given by:
[tex] Ka_{1} = \frac{[HSO_{3}^{-}][H_{2}CO_{3}]}{[HCO_{3}^{-}][H_{2}SO_{3}]} = 1.0\cdot 10^{-2} [/tex] (2)
This acid constant is related to the decomposition of H₂SO₃ to produce HSO₃⁻ and H₂CO₃.
From the second acid constant (H₂CO₃), we have:
[tex] Ka_{2} = \frac{[HCO_{3}^{-}][H_{2}SO_{3}]}{[HSO_{3}^{-}][H_{2}CO_{3}]} = 4.2 \cdot 10^{-7} [/tex] (3)
This second acid constant is for the decomposition of H₂CO₃ to produce H₂SO₃ and HCO₃⁻.
Since Ka₁ is bigger than Ka₂, the equilibrium of reaction (1) will shift to the right, to the production of HSO₃⁻ and H₂CO₃ (equation 2).
Therefore, the equilibrium will lie to the right side of the reaction.
Find more here:
- https://brainly.com/question/7145687?referrer=searchResults
- https://brainly.com/question/1619133?referrer=searchResults
I hope it helps you!
