KUNCI JAWABAN SOAL ULANGAN Ksp

1. Soal nomor 1


2. Soal nomor 2
b. Jumlah ion = 2 
s = √Ksp
s = √ 5 . 10-3
s = √ 50. 10-4
s = 7 . 10-2
atau :
AgBr ↔ Ag+ + Br-
s s s
Ksp = [Ag+] [Br-]
5 . 10-3 = s . s
s = √ 5 . 10-3
s = √ 50. 10-4
s = 7 . 10-2
c. Jumlah ion = 3

Atau
Ag2CO3 ↔ 2 Ag+ + CO3-2
s M 2s M s M
Ksp = [Ag+]2 [CO3-2]
Ksp = (2s)2 (s)
5.10-13 = 4s3

3. Soal nomor 3
Jumlah ion = 2 
s = √Ksp
s = √ 2,5 . 10-9
s = √ 25. 10-10
s = 5 . 10-5 M

atau :
CaCO3 ↔ Ca2+ + CO32-
s M s M s M
Ksp = [Ca2+] [CO32-]
5 . 10-3 = s . s
s = √ 2,5 . 10-9
s = √ 25. 10-10
s = 5 . 10-5 M


4. Soal nomor 4

Ag2CrO4 ↔ 2 Ag+ + CrO4-2
10-4 M 2.10-4 M 10-4 M
[Ag+] = 2.10-4 M
[CrO4-2] = 10-4 M

Atau
Ag2 CrO4 ↔ 2 Ag+ + CrO4-2
s M 2s M s M
Ksp = [Ag+]2 [CrO4-2]
Ksp = (2s)2 (s)
5.10-13 = 4s3


Ag2CrO4 ↔ 2 Ag+ + CrO4-2
10-4 M 2.10-4 M 10-4 M
[Ag+] = 2.10-4 M
[CrO4-2] = 10-4 M


5. Soal nomor 6

Jumlah ion = 3
Ksp = 4s3
Ksp = 4. (0,2 . 10-3)3
Ksp = 4 . 0,008 . 10-9
Ksp = 32 . 10-12
Atau
Ag2Cr2O7 ↔ 2 Ag+ + Cr2O7-2
0,2 . 10-3 M 0,4.10-3 M 0,2 . 10-3 M
Ksp = [Ag+]2 . [Cr2O7-2]
Ksp = (0,4.10-3)2 . (0,2 . 10-3)
Ksp = 16 . 10-8 . 2 . 10-4
Ksp = 32 . 10-12

6. Soal nomor 7
Ag2CrO4 ↔ 2 Ag+ + CrO4-2
10-4 M 2 . 10-4 M 10-4 M
Ksp = [Ag+]2 [CrO4-2]
Ksp = (2 . 10-4)2 (10-4)
Ksp = 4 . 10-12

AgNO3  Ag+ + NO3-
0,01 M 0,01 M
Ksp = [Ag+]2 [CrO4-2]
4 . 10-12 = (0,01)2 [CrO4-2]
[CrO4-2] = 4 . 10-12 / 10-4
[CrO4-2] = 4 . 10-8 M
Kelarutan Ag2CrO4 = 4 . 10-8 M


7. Soal nomor 14
a. NaCl  Na+ + Cl-
0,1 M 0,1 M
AgCl ↔ Ag+ + Cl-
Ksp = [Ag+] [Cl-]
2 . 10-10 = [Ag+] (0,1)
[Ag+] = 2 . 10-10 / (0,1)
[Ag+] = 2 . 10-9 M

b. CaCl2  Ca+2 + 2Cl-
0,1 M 0,2 M
AgCl ↔ Ag+ + Cl-
Ksp = [Ag+] [Cl-]
2 . 10-10 = [Ag+] (0,2)
[Ag+] = 2 . 10-10 / (0,2)
[Ag+] = 1. 10-9 M

8. Soal nomor 16
mol Ca+2 = (100/1000) . 0,3
mol Ca+2 = 0,03 mol
mol F- = (200/1000). 0,06
mol F- = 0,012 mol
volume total = 300 ml
[Ca+2] = 0,03/300
[Ca+2] = 10-4 M
[F-] = 0,012/300
[F-] = 4 . 10-5 M
CaF2 ↔ Ca+2 + 2F-
Q = [Ca+2] [F-]2 = (10-4) (4 . 10-5)2
[Ca+2] [F-]2 = 16 . 10-14
16 . 10-14 < Ksp
CaF2 tidak mengendap


9. Soal nomor 17
mol Pb+2 = (200/1000) (0,1)
mol Pb+2 = 0,02 mol
mol I-1 = (300/1000) (0,1)
mol I- = 0,03 mol
volum total = 500 ml
[Pb+2] = 0,02/500
[Pb+2] = 0,4 . 10-4 M
[I-] = 0,03/500
[I-] = 0,6 . 10-4 M
PbI2 ↔ Pb+2 + 2 I-1
Q = [Pb+2] [I-]2 = 0,4 . 10-4 (0,6 . 10-4)2
= 4.10-5 . 0,36. 10-8
= 144 . 10-15
144 . 10-15 < Ksp
PbI2 tidak mengendap

10. Soal nomor 20