Calculate the pH of a 0.020 M H2CO3 solution. At 25 °C, Ka1 = 4.3 × 10-7. H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO3-(aq)

Explanation:

H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)

Let d be the degree of dissociation

.02( 1-d )                              .02d          .02d

Dissociation constant Ka₁ is given

4.3 x 10⁻⁷  = .02d x .02d / .02( 1-d )

= .004 d² / .02  ( neglecting d in denominator )

= .02 d²

d² = 215 x 10⁻⁷

d = 4.636 x 10⁻³

= .004636

concentration of H₃O⁺

= d x .02

= .004636 x .02

= 9.272 x 10⁻⁵

pH = - log [ H₃O⁺ ]

= - log ( 9.272 x 10⁻⁵ )

5 - log 9.272

= 5 - .967

= 4.033 .

i believe the answer is a.

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