C6H5COOH(aq) + CH3COO^-(aq) ⇌ C6H5COO^-(aq) + CH3COOH(aq) is 3.6
The equilibrium constant for the reaction
C6H5COOH(aq) + CH3COO^-(aq) ⇌ C6H5COO^-(aq) + CH3COOH(aq)
is 3.6 at 25℃.
If Ka for CH3COOH is 1.8×10^(-5),
what is the acid dissociation constant for C6H5COOH?
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▶ 참고: 이론적 평형상수
[ https://ywpop.tistory.com/4154 ]
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C6H5COOH(aq) + CH3COO^-(aq) ⇌ C6H5COO^-(aq) + CH3COOH(aq)
K = [C6H5COO^-] [CH3COOH] / [C6H5COOH] [CH3COO^-]
CH3COOH(aq) ⇌ H^+(aq) + CH3COO^-(aq)
Ka1 = [H^+] [CH3COO^-] / [CH3COOH]
C6H5COOH(aq) ⇌ H^+(aq) + C6H5COO^-(aq)
Ka2 = [H^+] [C6H5COO^-] / [C6H5COOH]
H^+(aq) + CH3COO^-(aq) ⇌ CH3COOH(aq)
Ka3 = [CH3COOH] / [H^+] [CH3COO^-]
아세트산 이온화의 역반응이므로,
Ka3 = 1 / Ka1
Ka2 × Ka3 = K 이므로,
Ka2 = K / Ka3
= 3.6 / (1 / (1.8×10^(-5)))
= 6.5×10^(-5)
답: 6.5×10^(-5)
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