Respuesta :
When a solution of Pb(NO3)2(aq) is mixed with a solution of KI(aq), a precipitate of PbI₂ will form; K⁺ and NO₃⁻ are spectator ions.
Explanation:
When an aqueous solution of lead nitrate (Pb(NO₃)₂ is mixed with aqueous solution of potassium iodide (KI), then there is a precipitate formation of lead iodide (PbI₂), and the potassium (K⁺) ion and nitrate (NO₃⁻) ion acts as spectator ions that is ions do not involved in the reaction.
The reaction can be represented as,
Pb(NO₃)₂(aq) + 2 KI (aq) → PbI₂(s) + 2KNO₃(aq)
The ionic equation can be written as,
Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
On both sides of the equation, we have K⁺ and NO₃⁻ ions, which gets cancelled, and these 2 ions are called as spectator ions.
What will happen when Pb(NO3)2(aq) is mixed with a solution of KI(aq) is;
Option D; A precipitate of PbI2 will form; K+ and NO3– are spectator ions.
We want to find what will happen when Pb(NO3)2(aq) is mixed with a solution of KI(aq)
Let us write a balanced equation of this reaction.
Pb(NO3) 2 (aq) + 2KI(aq) = PbI2(s) + 2KNO3(aq)
To answer this question properly, we need to write an ionic equation of the balanced equation above.
Let us first list all the ions present which are;
Pb2+(aq) + 2NO3-(aq) and K+(aq) + I-(aq)
Writing the ionic equation now gives us;
Pb2+(aq) + 2NO3-(aq) + K+(aq) + I-(aq) = PbI2(s) + 2K+(aq) + 2NO3-(aq)
In this ionic equation, we see that it formed PbI2(s) which is a precipitate.
Now, the ions 2K+(aq) + 2NO3-(aq) are on the same side and as such are not participating in the reaction. Thus, they can be called spectator ions.
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