Answer:
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 .
