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mol Ba(OH)₂ = 100 x 0,1 = 10 mmol
Persamaan reaksi
2CH₃COOH + Ba(OH)₂ → Ba(CH₃COO)₂ + 2H₂O
a 20 10
r 20 10
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s - - 20 mmol
Karena pada akhir reaksi hanya terdapat garamnya saja, maka akan terbentuk sistem hidrolisis garam
Ba(CH₃COO)₂ ⇆ Ba²⁺ + 2CH₃OO⁻
20 mmol 20 40 mmol
M CH₃COO⁻ = mol/Volume campuran
M CH₃OO⁻ = 40/(200 + 100)
M CH₃COO⁻ = 0,133 M
[OH⁻] = √Kw/Ka • M anion garam
[OH⁻] = √Kw/Ka • M CH₃OO⁻
[OH⁻] = √10⁻¹⁴/10⁻⁵ • 0,133
[OH⁻] = √1,33 x 10⁻¹⁰
[OH⁻] = 1,15 x 10⁻⁵
pOH = -log [OH⁻]
pOH = -lig [1,15 x 10⁻⁵]
pOH = 5 - log 1,15
pH = 14 - pOH
pH = 14 - (5 - log 1,15)
pH = 9 + log 1,15
Jadi, pH larutan adalah 9 + log 1,15