chemistry (buffer solution)

CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq)

At equilibrium :
[CH3COOH] ≈ [CH3COOH]­o = 0.01 mol dm⁻³
[CH3COO⁻] = [CH3COONa]o= (5.2/82)/1 = 0.0634 mol dm⁻³

Ka = [CH3COO⁻][ H⁺]/[CH3­COOH]
pH = -logKa - log([CH3COOH]/[ CH3COO⁻])
pH = -log(1.74 x 10⁻⁵) - log(0.01/0.0634)
pH = 5.56

2012-11-16 22:42:03 補充：
CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq) .. Ka = 1.74 x 10⁻⁵

Initially :
[CH3COOH]o = 0.01 mol dm⁻³
[CH3COO⁻]o = [CH3COONa]o= (moles of CH3COONa)/volume = (5.2/82)/1 = 0.0634 mol dm⁻³

2012-11-16 22:42:21 補充：
As Ka is very small and also due to the common ion effect of the presence of CH3COO⁻ ions in the solution, the dissociation of CH3COOH is to a very small extent.
Hence, at equilibrium :
[CH3COOH] ≈ [CH3COOH]¬o = 0.01 mol dm⁻³
[CH3COO⁻] = [CH3COO⁻]o = 0.0634 mol dm⁻³

2012-11-16 22:43:03 補充：
Ka = [CH3COO⁻][ H⁺]/[CH3¬COOH]

Take logarithm on the both sides:
logKa = log([CH3COO⁻][ H⁺]/[CH3¬COOH])
logKa = log[CH3COO⁻] + log[ H⁺] - log[CH3¬COOH]
-log[ H⁺] = -logKa + log[CH3COO⁻] - log[CH3¬COOH]
pH = -logKa - log([CH3COOH]/[ CH3COO⁻])
pH = -log(1.74 x 10⁻⁵) - log(0.01/0.0634)
pH = 5.56

都係唔係幾明
可唔可以詳細少少
唔該