No 1-5 bantuin ya !!!
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Mr MgCl2 = 95
Massa MgCl2 = 7.6 mg = 7.6×10^-3 gram
mol MgCl2 = gram / Mr = 7.6 × 10^-3/ 95
= 8×10^-5 mol
[MgCl2] = mol / liter = 8×10^-5 / 0.1 = 8×10^-4 M
MgCl2 ---> Mg + 2Cl
s (2s)^2
Ksp = 4s^3
= 4 (8×10^-4)^3
= 4 (512 × 10^-12)
= 2048 × 10^-12
Ksp = 2.048 × 10^-9
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| No. 2|
pH = 11+log 4
pOH = 3-log 4 -----> [OH-] = 4×10^-3
Fe (OH)2 ----> Fe + 2 OH
Ksp = s s^2
4×10^-13 = s (4×10^-3)^2
4×10^-13 = 16×10^-6 s
s = 2.5 × 10^-8
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| No. 3|
NaCl ---> Na + Cl
0.02 0.02
Pb (No3)2 ---> Pb + 2 No3
0.2 0.2
PbCl2 ---> Pb + 2cl
Q = s s^2
Q = 2×10^-1 . (2×10^-2)^2
= 2×10^-1 . 4×10^-4
Q = 8×10^-5
--> Ksp PbCl2 = 1.7 × 10^-5
Q > Ksp --> Maka terjadi endapan
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|No. 4|
Ag2CrO4 ---> 2 Ag + CrO4
2s s
Ksp = 4s^3
s = (akar pangkat tiga Ksp dibagi 4)
s = 8.43 × 10^-5
Na2CrO4 ---> 2Na + Cro4
0.1 0.1
Ag2Cro4 ----> 2 Ag + Cro4
Ksp = s^2 s
2.4×10^-12 = s^2 10^-1
s = ( akar dari 2.4×10^-11)
s = 4.89×10^-5
Kesimpulan :
Penambahan Na2CrO4 Menurunkan
kelarutan Ag2CrO4
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| No.5|
Mg (OH)2 ----> Mg + 2 OH
s 2s
Ksp = 4s^3
s = (akar pangkat tiga Ksp dibagi 4)
s = 6.69 × 10^-5
pH = 9
pOH = 5 -----> [OH-] = 10^-5
Mg (OH)2 ----> Mg + 2 OH
Ksp = s s^2
1.2 × 10^-12 = s . (10^-5)^2
s = 1.2 × 10^-12/ 10^-10
s = 1.2 × 10^-2