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pH Ba(OH)₂ = 13
pOH = 1
[OH⁻] = 10⁻¹
[OH⁻] = M x Valensi
10⁻¹ = M x 2
M = 0,05 M
* Mencari mol Ba(OH)₂ dan HCN
mol Ba(OH)₂ = 100 x 0,05 = 5 mmol
mol HCN = 100 x 0,1 = 10 mmol
Persamaan reaksi :
2HCN + Ba(OH)₂ → Ba(CN)₂ + 2H₂O
a 10 5
r 10 5
____________________________________
s - - 5 mmol
Terbentuk sistem Hidrolisis Garam
* Mencari mol anion garam (CN⁻)
Ba(CN)₂ ⇆ Ba²⁺ + 2CN⁻
5 5 10 mmol
mol CN⁻ = 10 mmol
* Mencari Molaritas CN⁻
M = mol/Volume total
M = 10/200
M = 0,05 M
* Mencari [OH⁻]
[OH⁻] = √Kw/Ka • M Garam
[OH⁻] = √10⁻¹⁴/4,9 x 10⁻¹° • 0,05
[OH⁻] = √1,02 x 10⁻⁶
[OH⁻] = 10⁻³
* Mencari pOH
pOH = -log [OH⁻]
pOH = -log [10⁻³]
pOH = 3
* Mencari pH
pH = 14 - pOH
pH = 14 - 3
pH = 11
Jadi, pH larutan adaah 11