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kozerog [31]
3 years ago
9

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.

Chemistry
1 answer:
Aneli [31]3 years ago
5 0
Answer is: Ka for phenol (C₆H₅OH) is 2.67·10⁻¹¹.
<span>
Chemical reaction: C</span>₆H₅OH(aq) + H₂O(l) → C₆H₅O⁻(aq) + H₃O⁺(aq) or
C₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺<span>(aq).
</span>Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH<span>]. 
</span>[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.<span>

</span>
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