If the [h+] of a 0.205m solution of phenol (c6h5oh) at 25ºc is 2.340 10-6, what is the ka for phenol? phenol is monoprotic.

Answer is: Ka for phenol (C₆H₅OH) is 2.67·10⁻¹¹.

Chemical reaction: C₆H₅OH(aq) + H₂O(l) → C₆H₅O⁻(aq) + H₃O⁺(aq) or
C₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺(aq).
Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH]. 
[H⁺] = [C₆H₅O⁻] = 2.340·10⁻⁶ M; equilibrium concentration.
[C₆H₅OH] = 0.205 M - 2.340·10⁻⁶ M.
[C₆H₅OH] = 0.20499 M.
Ka = (2.340·10⁻⁶ M)² ÷ 0.20499 M.
Ka = 2.67·10⁻¹¹.

Answer) by studying proton collisions hope this : )