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Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in the complete ionic equation?

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Answer:

The question is incomplete and confusing.

  • In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

  • In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

1. Write the balanced molecular equation:

  • Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

2. Write the ionizations for the ionic aqueous compounds:

  • Pb(NO₃)₂(aq) →  Pb⁺²(aq) + 2NO₃⁻(aq)

  • 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)

  • 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)

3. Write the complete ionic equation:

Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) +  2K⁺(aq) + 2NO₃⁻(aq)

Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.

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