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`        	The solubility of Mg(OH)2  in pure water is 9.57 x 10^-3 g L^-1  . Its solubility (in g L^-1 ) in 0.02 moles Mg(NO3)2 will be`
2 years ago

nigamananda mishra
25 Points
```							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)3 = 17.96 x 10-12 Suppose, the solubility of Mg(OH)2 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
```
2 years ago
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