Page 110 - Pharmaceutical_Analytical_Chemistry_1_Theoretical_Notes_Level_1

Mansoura National University
Pharm D-Clinical Pharmacy Program Level 1 Pharm. Anal. Chem. 1 (PC 101)

-3
Q2: The solubility of CaF is 2.3 x 10 g/100ml, calculate Ksp (Mol.wt. 78 g/mol).
2
-3
We must first convert the solubility of calcium fluoride from 2.3 x 10 g/100mL
to moles per liter.
-3 -2
The solubility of CaF = 2.3 x 10 g/100 ml = 2.3 x 10 g/1000 ml
2

No. of moles = Weight / Mol.wt
No. of moles per litre=Molar Concentration

2.3  10 -2
Molar solubility = = 9 . 2  10 − 4 M
78

Q3: The solubility of CaF 2 is 2.3 x 10 g/100ml, calculate Ksp (Mol.wt. 78 g/mol).
-3

2.3  10 -2
Molar solubility = = 9 . 2  10 − 4 M
78

2+
-
CaF 2 ↔ Ca + 2F
1 mole 1 mole 2 mole
-4
-4
-4
2.9 x 10 2.9 x 10 2(2.9 x 10 )
- 2
2+
K sp = [Ca ] [F ]
-4
2
-4
-11
K sp = ( 2.9 x 10 ) [2(2.9 x 10 )] = 9.7 x 10
Calculating the Solubility from Ksp
-10
Q1: What is the molar solubility of AgCl in water at 25 °C ? Ksp (AgCl) = 1.7 x 10
Solution:

✓ Write out the balanced chemical equation for the substance
✓ Write the K sp expression

✓ Solve for x
-
+
AgCl ↔ Ag + Cl
X X X

-
2
+
-10
K sp = [Ag ] [Cl ] = (X) (X) = X = 1.7 x 10
-5
Molar solubility, X = 1.3 x 10 M
105