1) Larutan jenuh basa L(OH)3 mempunyai pH = 10. Tetapan hasil kali kelarutan (Ksp) basa tersebut adalah... A. 3,3 x 10^-17 B. 4,0 x 10^-16 C. 2,7 x 10^-15 D. 4,0 x 10^-12 E. 3,3 x 10^-5
2) Ksp X(OH)2 adalah 4 x 10^-12 maka pH larutan jenuh X(OH)2 adalah... A. 9 B. 9 + log2 C. 10 D. 10 + log2 E. 11 + log2
Nomor 2
pOH = 14 - 10 = 4
[OH-] = 10-4
L(OH)3 -> L3+ + 3OH-
[L3+] = 1/3 x 10-4 = 3,3 x 10-5
Ksp = [L3+] [OH-]3
= (3,3 x 10-5) (10-4)3
Ksp = 3,3 x 10-17 (a)
2. X(OH)2 -> X2+ + 2OH-
Ksp = 4s3
s = 3akar Ksp / 4
s = 3akar 4 x 10-12 / 4
s = 3akar 1 x 10 -12
s = 1 x 10-4
[OH-] = 2s
= 2(1 x 10-4)
= 2 x 10-4
pOH = 4 - log 2
pH = 14 - (4 - log 2)
pH = 10 + log 2 (d)