0.674 M HF 용액의 pH

Calculate the pH of the weak acid HF at equilibrium,

if the initial concentration of HF was 0.674 M.

Ka = 2.39×10^(-5)

HF(aq) ⇌ H^+(aq) + F^-(aq)

Ka = [H^+] [F^-] / [HF]

= (x) (x) / (C–x)

= x^2 / (0.674–x)

0.674–x ≒ 0.674 라 근사처리하면,

Ka = x^2 / 0.674

x = [Ka × 0.674]^(1/2)

= [2.39×10^(-5) × 0.674]^(1/2)

= 0.00401355 M = [H^+]

0.674 M × (5/100) = 0.0337 M > 0.00401355 M

---> 근사처리 ok.

pH = –log[H^+]

= –log(0.00401355)

= 2.39647

답: pH = 2.40

[ 관련 글 https://ywpop.tistory.com/19715 ]

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