Neet Solved Paper 2015 Question 26
Question: The $ K _{sp} $ of $ Ag _2CrO _4,\ AgCl,\ AgBr $ and $ AgI $ are respectively, $ 1.1\times {10^{-12}},1.8\times {10^{-10}} $ $ 5.0\times {10^{-13}},8.3\times {10^{-17}} $ . Which one of the following salts will precipitate last if $ AgNO _3 $ solution is added to the solution containing equal moles of $ NaCl,NaBr,NaI $ and $ Na _2CrCO _4 $ ?
Options:
A) $ Agl $
B) $ AgCl $
C) $ AgBr $
D) $ Ag _2CrO _4 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ Ag _2CrO _4\to 2A{g^{+}}+CrO_4^{2-} $
Solubility product $ K _{sp}={{(2s)}^{2}}\times s=4s^{3} $ $ K _{sp}=(1.1\times {10^{-12}}) $ $ S=\sqrt[3]{\frac{K _{sp}}{4}},=0.65\times {10^{-4}} $
$ AgCl\to A{g^{+}}+C{l^{-}} $ $ K _{sp}=S\times S $ $ (K _{sp}=1.8\times {10^{-10}}) $ $ S=\sqrt{K _{sp}},=1.34\times {10^{-5}} $
$ AgBr\to A{g^{+}}+B{r^{-}} $ $ K _{sp}=S\times S $ $ (K _{sp}=5\times {10^{-13}}) $ $ S=\sqrt{K _{sp}}=0.71\times {10^{-6}} $
$ AgI \to A{g^{+}}+{I^{-}} $ $ K _{sp}=S\times S $ $ (K _{sp}=8.3\times {10^{-17}}) $ $ S=\sqrt{K _{sp}}=0.9\times {10^{-8}} $
$ \because $ Solubility of $ Ag _2CrO _4 $ is highest. So, it will precipitate last.
BETA
