Q 15) Lead nitrate is added to a solution containing potassium iodide.
a) Write the name and colour of the compound precipitated.
b) Write the balanced chemical reaction involved?
c) Name the type of reaction.
Answers
Answer:
2KI(aq)+Pb(NO
3
)
2
(aq)→PbI
2
(s)+2KNO
3
In the above double displacement reaction, potassium Iodide(KI) and lead nitrate (Pb(NO
3
)
2
) dissociate in their aqueous states to form ions. The lead (Pb
2+
) ions combine with the iodide (I
−
) ions to form precipitates of lead iodide (PbI
2
). If Lead sulphate is used in place of lead nitrate, no precipitates of lead iodide will be formed, because lead sulphate being insoluble in water does not give Pb
2+
ions which can combine with I
−
to form PbI
2
. On the other hand, lead acetate dissociates in aqueous state to give Pb
2+
ions and CH
3
COO
−
ions . Therefore potassium iodide combines with lead acetate to form precipitates of lead iodide and potassium acetate.
2KI(aq)+Pb(CH
3
COO)
2
(aq)→PbI
2
(s)+2CH
3
COOK
Explanation:
Answer:
a) KI2 (Lead Iodide) which is Yellow in colour
b) Pb(NO3)2 + 2KI ---> PbI2 + 2KNO3
c) Double displacement/ precipitation reaction
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