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Balance the following
solution of lead(II) nitrate and potassium chromate yield solid lead(II) chromate in a potassium nitrate solution
Solid sodium hydroxide is placed into sulfuric acid resulting in aqueous sodium sulfate and water
Hydrogen chloride gas reacts with ammonia gas to produce solid ammonium chloride
Hydrobromic acid can be neutralized by adding aqueous lithium hydroxide to produce aqueous lithium bromide and water
Carbonic acid is an unstable acid that breaks down to produce water and carbon dioxide gas
Answers
In this case, lead(II) nitrate, Pb(NO3)2, and sodium iodide, NaI, both soluble in water, will exist as ions in aqueous solution
Pb(NO3)2(aq]→Pb2+(aq]+2NO−3(aq]
NaI(aq]→Na+(aq]+I−(aq]
When these two solutions are mixed, the lead(II) cations, Pb2+, and the iodide anions, I−, will bind to each other and form lead(II) iodide, an insoluble ionic compound.
The other product of the reaction is aqueous sodium nitrate, NaNO3, which will exist as ions in solution.
You can thus say that
Pb(NO3)2(aq]+2NaI(aq]→PbI2(s]⏐⏐↓+2NaNO3(aq]
The complete ionic equation, which includes all the ions present in solution, will look like this
Pb2+(aq]+2NO−3(aq]+2Na+(aq]+2I−(aq]→PbI2(s]⏐⏐↓+2Na+(aq]+2NO−3(aq]
The net ionic equation, which eliminates spectator ions, i.e. the ions that are present on both sides of the equation, will look like this
Pb2+(aq]+2NO−3(aq]+2Na+(aq]+2I−(aq]→PbI2(s]⏐⏐ ⏐↓+2Na+(aq]+2NO−3(aq]
This will be equivalent to
Pb2+(aq]+2I−(aq]→PbI2(s]⏐⏐↓
Lead(II) iodide is a yellow solid that precipitates out of solution.
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Explanation:
You're dealing with a double replacement reaction in which two soluble ionic compounds in aqueous solution react to form an insoluble solid that precipitates out of solution.
In this case, lead(II) nitrate,
Pb
(
NO
3
)
2
, and sodium iodide,
NaI
, both soluble in water, will exist as ions in aqueous solution
Pb
(
NO
3
)
2(aq]
→
Pb
2
+
(aq]
+
2
NO
−
3(aq]
NaI
(aq]
→
Na
+
(aq]
+
I
−
(aq]
When these two solutions are mixed, the lead(II) cations,
Pb
2
+
, and the iodide anions,
I
−
, will bind to each other and form lead(II) iodide, an insoluble ionic compound.
The other product of the reaction is aqueous sodium nitrate,
NaNO
3
, which will exist as ions in solution.
You can thus say that
Pb
(
NO
3
)
2(aq]
+
2
NaI
(aq]
→
PbI
2(s]
⏐
⏐
↓
+
2
NaNO
3(aq]
The complete ionic equation, which includes all the ions present in solution, will look like this
Pb
2
+
(aq]
+
2
NO
−
3(aq]
+
2
Na
+
(aq]
+
2
I
−
(aq]
→
PbI
2(s]
⏐
⏐
↓
+
2
Na
+
(aq]
+
2
NO
−
3(aq]
The net ionic equation, which eliminates spectator ions, i.e. the ions that are present on both sides of the equation, will look like this
Pb
2
+
(aq]
+
2
NO
−
3(aq]
+
2
Na
+
(aq]
+
2
I
−
(aq]
→
PbI
2(s]
⏐
⏐
⏐
↓
+
2
Na
+
(aq]
+
2
NO
−
3(aq]
This will be equivalent to
Pb
2
+
(aq]
+
2
I
−
(aq]
→
PbI
2(s]
⏐
⏐
↓
Lead(II) iodide is a yellow solid that precipitates out of solution.