Question

2.hitunglah Ksp dari larutan Ca(OH)2 yang memiliki pH sebesar = 9+log 3

Answer 1

Diketahui :
pH = 9 + log 3

Ditanya : Ksp Ca(OH)2 ?

Jawab :

pOH :
= 14 - pH
= 14 - ( 9 + log 3)
= 14 - 9 - log 3
= 5 - log 3

[OH-] :
- log [OH-] = pOH
- log [OH-] = 5 - log 3
- log [OH-] = - log 3 × 10^-5
[OH-] = 3 × 10^-5


Reaksi :
Ca(OH)2 <---> Ca²+ + 2OH-
____s___________s______2s

Kelarutan Ca(OH)2 (s) :
= ½ × Konsentrasi OH-
= ½ × 3 × 10^-5
= 1,5 × 10^-5

Ksp :
= [Ca²+] [OH-]²
= s . (2s)²
= s . 4s²
= 4s³
= 4 ( 1,5 × 10^-5)³
= 4 × 3,375 × 10^-15
= 1,35 × 10^-14


Catatan :
( ^ ) : Pangkat