Q:-3 When you have mixed the solution of lead (11) nitrate and potassium iodode.
a) What was the colour of the precipitate ?
b) Write the chemical equationfor the reaction.
c) Is this a double displacement reaction.
Answers
Answer:
it this a double displasement reation
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