A solution X on mixing with AgNO₃ solution, gives a white precipitate soluble in NH₃₍ₐԛ₎. A solution Y, when added to X, also gives a white precipitate which is soluble on boiling. Solution Y contains
A. Ag⁺ ion
B. Pb²⁺ion
C. Pb⁴⁺ion
D. Zn²⁺ion
QUICK ANSWER…
B
DETAILS…
Identifying X is key to finding the possible identity of Y.
Which compound of silver is insoluble? Most silver halides (except silver fluoride), are insoluble, so X is most likely hydrochloric acid (HCl).
When HCl is mixed with silver nitrate, a white precipitate of silver chloride is formed.
When this precipitate of silver chloride is treated with ammonia, the precipitate dissolves into soluble ammonium chloride and silver nitrate according to the equation:
AgNO₃ + HCl = AgCl + HNO₃
AgCl + HNO₃ + NH₃ = NH₄CL + AgNO₃
So, X has been identified as HCl.
Next is to identify the ion that will also form a precipitate with HCl, lead (ii) ion is the ion,
When HCl reacts with lead (ii) salt, lead (ii) chloride will be formed. Now lead (ii) chloride is insoluble in cold water but soluble in hot water.
Now for the right answer to the above question:
- Option A is incorrect. silver chloride (which will be formed) is not soluble on boiling.
- Option B is correct. lead (ii) chloride is insoluble in cold water but soluble in hot water.
- C is incorrect. lead (iv) ion will also give lead (ii) chloride but with chlorine gas which on boiling will not dissolve.
- D is not the correct answer. Zinc chloride, which is expected to be formed is very soluble in water, and so does not need heat.
KEY-POINTS…
You may please note these/this:
- All silver halides (except silver fluoride) are insoluble in water.
- Lead (ii) chloride is insoluble in cold water, but soluble in hot water.
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/ culled from 2015 JAMB-UTME Chemistry past question 3 /
