Based on the calculations, the pH of a 0.47 M solution of carbonic acid (H₂CO₃) is equal to 3.35.
First of all, we would write a properly balanced chemical equation for these chemical reactions:
H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4.3 × 10⁻⁷.
For the second chemical reaction:
HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5.6 × 10⁻¹¹.
Next, we set up an equation from the ICE table;
[HCO₃⁻] = [H⁺] = x.
[H₂CO₃] = 0.47 - x.
Ka₁ = ([HCO₃⁻] × [H⁺])/[H₂CO₃].
4.3 × 10⁻⁷ = (x × x)/(0.47 - x).
4.3 × 10⁻⁷ = x²/(0.47 - x).
x² = 0.0000002021 - 4.3 × 10⁻⁷x
x² + 4.3 × 10⁻⁷x - 0.0000002021 = 0
By solving the quadratic equation, we have:
x = [H⁺] =0.000449 M.
Now, we can determine the pH by using this formula;
pH = -log[H⁺]
pH = -log(0.000449 M).
pH = -(-3.35)
pH = 3.35.
Note: The second dissociation from Ka₂ would be so small, so it's considered negligible.
Read more on pH here: brainly.com/question/24233266
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