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Arrange in increasing order of solubility of AgBr in solutions given here
(i) 0.1 M (ii) 0.1 M (iii) 0.2 M NaBr (iv) pure water
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(i) 0.1 M (ii) 0.1 M (iii) 0.2 M NaBr (iv) pure water
Solution:
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Verified Answer
The correct answer is:
(iii) < (ii) < (iv) < (i)
AgBr in (Solubility increases, due to stable complex)
in 0.1 M (Solubility decreases)
in 0.2 M NaBr (Solubility decreases more)
in pure water (Normal dissociation)
Hence the order is
(iii) < (ii) < (iv) < (i)
in 0.1 M (Solubility decreases)
in 0.2 M NaBr (Solubility decreases more)
in pure water (Normal dissociation)
Hence the order is
(iii) < (ii) < (iv) < (i)
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