A yellow solid is formed when aq KI is mixed with a transparent solution. Identify the transparent solution, the yellow solid and write reaction involved in this.
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:
hii friend are you also in class 10
When two substances (can be aqueous or liquids) in solution react, they yield a solid and an aqueous compound.
This is a double displacement reaction in which the solid is known as a precipitate.
You can determine which compound is the precipitate by using a solubility chart.
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Soluble =>
a
q
compound.
Low solubility => precipitate will form
(
s
)
.
An example of a yellow precipitate forming is through the reaction of potassium iodide,
K
I
with lead (II) nitrate,
P
b
(
N
O
3
)
2
.
The products are lead (II) iodide,
P
b
I
2
, and potassium nitrate,
(
K
N
O
3
)
.