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Question:
The solubility product of Al(OH) 3 is 2.7 × 10-11. Calculate its solubility in gL-1 and also find out the pH of this solution. (Atomic mass of Al = 27 u).
Solution:
The equation of disassociation of Al(OH)3will be-
Al(OH)3 ⇌ Al3+ + 3OH-
We know that,
Ksp= [Al3+] [OH-] 3-
=(s) × (3s) 3 =27s4
S4 = Ksp/27
= 2.7x 10-11/27
s4= 10-12
s= (10-12)1/4 =10-3 mol/L
Now, molar mass of Al(OH)3 =78
Solubility= molar mass ×s
= 78 ×10-3
= 7.8 × 10-2 g/L
NOW, we know that-
pH = 14 – pOH
[OH]= 3s = 3 × 10-3
pOH= 3-log3
pH = 14 – 3 + log3
= 11.4771
Hence the pH of the solution will be 11.4771 and solubility in g/L will be 7.8×10-2 g/L.
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