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# The solubility of Mg(OH)2 in pure water is 9.57*10-3 g/L. Calculate its solubility (in gram per litre) in 0.02M Mg(NO3)2 solution.

Asked by Nidhi Jarodiya 2nd October 2013, 3:54 PM
solubility of Mg(OH)2 in pure water = 9.57 x 10-3 g/L
= 9.57 x 10-3   /molar mass  mole/L
=  9.57 x 10-3   /58  mole/L = 1.65 x 10-4 mol/L

Ks for Mg(OH)2 = s x (2s)2 = 4s3
= 4 x (1.65 x 10-4 mol/L)
= 17.96 x 10-12

Suppose, the solubility of Mg(OH)in the presence of  Mg(NO3)2  = s'

So, Mg2+ = s'+c = s' + 0.02
[OH'] = 2s'

Therefore, Ks = (s' + 0.02) (2s')2

17.96 x 10-12 =  4 (s' + 0.02) (s')2

or, 17.96 x 10-12/4 = s'3 + 0.02 (s' )2 , on neglecting s'3

4.4921 x 10-12 = 0.02 (s' )2

(s' )2 = 14.98 x 10-6 mol/l

Solubility in g/l
= 14.98 x 10-6 x58 = 8.69 x 10-4 g/l
Answered by Expert 3rd October 2013, 9:21 AM
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