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= 0,15 x 0,1
= 0,015 mol
mol NaOH = Molaritas x Volume
= 0,1 x 0,15
= 0,015 mol
Persamaan reaksi :
CH₃OOH + NaOH → CH₃COONa + H₂O
a 0,015 0,015
r 0,015 0,015
_____________________________________
s - - 0,015
Karena antara asam lemah dan basa kuat habis bereaksi, maka terbentuk sistem hodrolisis garam.
M garam = mol/Volume total
M garam = 0,015/(0,1 + 0,15)
M garam = 0,015/0,25
M garam = 0,06 M
[OH⁻] = √(Kw/Ka) • M garam
[OH⁻] = √(10⁻¹⁴/10⁻⁵) • 0,06
[OH⁻] = √6 x 10⁻¹¹
[OH⁻] = 7,75 x 10⁻⁶
pOH = -log [OH⁻]
pOH = -log [7,75 x 10⁻⁶]
pOH = 6 - log 7,75
pH = 14 - pOH
pH = 14 - (6 - log 7,75)
pH = 8 + log 7,75
Jadi, pH campuran itu adalah 8 + log 7,75