Analytical Chemistry Question 186
Question: A solution containing $ A{s^{3+}},C{d^{2+}},N{i^{2+}} $ and $ Z{n^{2+}} $ is made alkaline with dilute $ NH_4OH $ and treated with $ H_2S $ . The precipitate obtained will consist of
Options:
A) $ As_2S_3 $ and CdS
B) CdS.NiS and ZnS
C) NiS and ZnS
D) Sulfide of all ions
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Answer:
Correct Answer: D
Solution:
$ Cd^{2+} $ and $ Zn^{2+} $ are the radicals of group II, whereas $ Ni^{2+} $ and $ Pb^{2+} $ are the radicals of group IV. The solubility product of group IV radicals is higher as compared to group II. $ NH_4OH $ increases the ionisation of $ H_2S $ by removing $ {H^{+}} $ of $ H_2S $ as unionisable water. $ H_2S\rightarrow 2{H^{+}}+{S^{2-}}; $
$ {H^{+}}+O{H^{-}}\xrightarrow{{}}H_2O $ Thus excess of sulphide ions are present which leads to the precipitation of all the four ions. Note: $ HCl $ decreases ionisjation of $ H_2S $ whereas $ NH_4OH $ increases the ionisation of $ H_2S $ .
BETA
