Answer:
The pH in a solution prepared by continuously bubbling SO₂ is 1.03.
Explanation:
Searching in google the reaction is:
SO₂(g) + H₂O(l) ⇄ H₂SO₃(ac) K = 1.33
The constant is:
[tex]K = \frac{[H_{2}SO_{3}]}{P_{SO_{2}}} = 1.33[/tex]
Having that K = 1.33 and P(SO₂) = 0.500 atm, the concentration of H₂SO₃ is:
[tex] [H_{2}SO_{3}] = K*P_{SO_{2}} = 1.33*0.500= 0.665 M [/tex]
Now, we have the following dissociation of H₂SO₃ in water:
H₂SO₃(ac) + H₂O(l) ⇄ HSO₃⁻(ac) + H₃O⁺(ac) Ka₁ = 1.5x10⁻²
0.665-x x x
[tex]Ka_{1} = \frac{[HSO_{3}^{-}][H_{3}O^{+}]}{[H_{2}SO_{3}]} = 1.5\cdot 10^{-2} [/tex]
[tex] 1.5\cdot 10^{-2} = \frac{x^{2}}{0.665 - x} [/tex]
Solving the above equation for x we have:
x = 0.093 = [H₃O⁺] = [HSO₃⁻]
Also, the HSO₃⁻ dissociates in water as follows:
HSO₃⁻(c) + H₂O(l) ⇄ SO₃²⁻(ac) + H₃O⁺(ac) Ka₂ = 6.3x10⁻⁸
0.093-y y 0.093+y
[tex]Ka_{2} = \frac{[SO_{3}^{2-}][H_{3}O^{+}]}{[HSO_{3}^{-}]} = 6.3\cdot 10^{-8} [/tex]
[tex] 6.3\cdot 10^{-8} = \frac{y(0.093 + y)}{(0.093 - y)} [/tex]
[tex] 6.3\cdot 10^{-8}*(0.093 - y) - y(0.093 + y) = 0 [/tex]
Solving the above equation for y we have:
y = 6.3x10⁻⁸
So, the concentration of H₃O⁺ is:
[tex] [H_{3}O^{+}] = 0.093 M + 6.3\cdot 10^{-8} M = 0.093 M [/tex]
Finally, the pH is:
[tex] pH = -log[H_{3}O{+}] = -log(0.093) = 1.03 [/tex]
Therefore, the pH in a solution prepared by continuously bubbling SO₂ is 1.03.
I hope it helps you!
