幫我證明這個化學公式

應是：pH = pKa - log([CH3­COOH]/[CH3COO⁻])
或者：pH = pKa + log([CH3­COO⁻]/[CH3COOH])


證明：
CH­3­COO⁻ + H2O ⇌ CH3COOH + OH⁻ (Kb)
Kb = [CH3COOH][OH⁻]/[CH3COO⁻]

KaKb = 10⁻¹⁴
KaKb = [H⁺][OH⁻]
Ka = [H⁺][OH⁻] x (1/Kb)
Ka = [H⁺][OH⁻] x [CH3COO⁻]/[CH3COOH][OH⁻]
Ka = [H⁺][CH3COO⁻]/[CH3COOH]
log(Ka) = log([H⁺][CH3COO⁻]/[CH3COOH])
log(Ka) = log[H⁺] + log([CH3COO⁻]/[CH3COOH])
log(Ka) = log[H⁺] + log([CH3COO⁻]/[CH3COOH])
-log[H⁺] = -log(Ka) + log([CH3COO⁻]/[CH3COOH])

pH = -log[H⁺] 及 p(Ka) = -log(Ka)
pH = pKa + log([CH3COO⁻]/[CH3COOH])

pH = pKa + log([CH3COOH]/[CH3COO⁻])⁻¹
pH = pKa - log([CH3COOH]/[CH3COO⁻])

同學，這應該是妳化學期末考的加分題吧

加分題有跟沒有都一樣拉

沒能力就不要加嘛...

可是俗話說...〝不齒下問〞

妳也沒錯拉

1/2離你不遠了

說錯...○培是2/3

您的證明從HA-->H+ + A- 就可以證明出來