Application of Statistical and mathematical equations in Chemistry Part 4 Solubility Theory Precipitation Percentage calculation

2. Chapter 4

4. Problem1
•Example:
•Find the solubility of 0.038g PbSO4 in liter water
•PbSO4(s) Pb+2
(aq) + SO4
-2
(aq) .
Q3-g
• Example: Find the solubility of Mg(OH)2
(s) in water
• Mg(OH)2
(s) 1 Mg+2
(aq) + 2 OH-1
(aq)
• while
• solubility Mg(OH)2=1.4*10^-4.
•

8. Problem1
•Example (1): Find the concentration of NH4
+ that needed to prevent
precipitation of Mg(OH)2 in liter of solution that contain 0.01 mole
NH3 and 0.001mole Mg+2 .
•(Ksp for Mg(OH)2 = 1.12 * 10-11 , K for NH3 = 1.8 * 10-5) .
•Example (2): Find the Ag+ concentration that needed to precipitate AgCl in
2.5 * 10-4M NaCl solution .
•(Ksp for AgCl = 2.8 * 10-8) .
•Example (Q/10): Calculate the solubility of CaF2 & SrF2 in the same
time , if you know the .
•(Ksp for both solute is = 4.0*10 ^-11 & 2.8 * 10-9) .

10. As a result of the majority of F ions coming from SrF 2 , the calculation of solubility
of SrF2 can be calculate as
[F]=X & [Sr]=X
Ksp= [F]^2*[Sr]= 2X^2* X= 4X^3
X= (2.8*10^-9/4)^1/3
X= 8.88*10^-4
The solubility of CaF2 is depending on F Conce coming from SrF2
[Ca++]= Ksp/[F-] = ( 4.0*10^-11)/2*8.88*10^-4
[Ca++]= 0.000012M

12. Problem1
•Example:
•Find the %Mn2O3 and %Mn in 3.04g of the sample that produce
0.252g of Mn3O4 .

15. 1 mole of KBr gives 1mole of AgBr
So
119.09 mg KBr gives 187.78mg of AgBr
While the amount of AgBr that coming out equile 804.3 mg
X mg KBr gives 804.3mg AgBr
X mg KBr = 804.3mg AgBr * 119.01 mg KBr)/ 187.78mg AgBr = 509.74 mg KBr
So , t