To determine the products of the reaction 2KI(aq) + Pb(NO3)2(aq), we first need to identify the ions present in the reactants and then consider the solubility rules to predict the state of the products.
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The reactants dissociate in aqueous solution as follows:
- KI dissociates into K⁺ and I⁻ ions.
- Pb(NO3)2 dissociates into Pb²⁺ and NO3⁻ ions.
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Now, we can predict the possible products when these ions combine:
- K⁺ ions will pair with NO3⁻ ions to form KNO3.
- Pb²⁺ ions will pair with I⁻ ions to form PbI2.
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The balanced reaction is: \[ 2KI(aq) + Pb(NO3)2(aq) \rightarrow 2KNO3(aq) + PbI2(s) \]
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Next, we determine the states of matter for the products:
- KNO3 is soluble in water, so it remains as an aqueous solution (KNO3(aq)).
- PbI2 is not soluble according to the solubility rules (specifically, most iodides are soluble except for those of lead, silver, and mercury), so it will precipitate out of the solution and is solid (PbI2(s)).
So the complete equation with states is: \[ 2KI(aq) + Pb(NO3)2(aq) \rightarrow 2KNO3(aq) + PbI2(s) \]
Finally, regarding the question of whether a precipitate will form: Yes, a precipitate of PbI2 will form in this reaction.