NCERT Class XI Chemistry Equilibrium Solutions
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Question : 67
Total: 73
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in table. Determine also the molarities of individual ions.
| Salt | Ksp value |
|---|---|
| Silver chromate | 1.1 × |
| Barium chromate | 1.2 × |
| Ferric hydroxide | 1.0 × |
| Lead chloride | 1.6 × |
| Mercurous iodide | 4.5 × |
Solution:
Silver chromate, A g 2 C r O 4
A g 2 C r O 4 ( s ) 2 A g ( a q ) + + C r O 2 – 4 ( a q )
K s p 4 S 3 10 – 12 10 – 5
∴[ A g + ] 10 – 5 10 – 4 M 10 – 5 M
Barium chromate,B a C r O 4
B a C r O 4 ( s ) B a ( a q ) 2 + + C r O 2 – 4 ( a q )
K s p S 2 10 – 10 10 – 5 M
[ B a 2 + ] [ C r O 4 2 – ] 10 – 5 M
Ferric hydroxide,F e ( O H ) 3
F e ( O H ) 3 ( s ) F e ( a q ) 3 + + 3 O H ( a q ) –
K s p 27 S 4 10 – 38 10 – 10 M
[ F e 3 + ] 10 – 10 M [ O H – ] 10 – 10 10 – 10 M
Lead chloride,P b C l 2
P b C l 2 ( s ) P b ( a q ) 2 + + 2 C l ( a q ) –
K s p 4 S 3 10 – 5
∴ S = 1.59 ×10 – 2 M
[ P b 2 + ] 10 – 2 M [ C l – ] 10 – 2 10 – 2 M
Mercurous iodide,H g 2 I 2
H g 2 I 2 ( s ) H g 2 ( a q ) 2 + + 2 I ( a q ) –
K s p 4 S 3 10 – 29 10 – 10 M
[ H g 2 2 + ] 10 – 10 M [ I – ] 10 – 10 10 – 10 M
∴
Barium chromate,
Ferric hydroxide,
Lead chloride,
∴ S = 1.59 ×
Mercurous iodide,
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