Understanding Ionic Equations: Pb2+ & I- Reactants

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Discussion Overview

The discussion revolves around understanding ionic equations, specifically focusing on the reactants Pb2+ and I- in the context of a precipitation reaction involving lead(II) iodide. Participants explore the roles of various ions in solution and the criteria for determining which ions are involved in the reaction.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • One participant states that an ionic equation shows the chemical species that participate in a reaction and questions the involvement of K+ and NO3- in the reaction between Pb(NO3)2 and KI.
  • Another participant suggests that since KNO3 and KI are soluble, the ions that do not precipitate may not be actively participating in the reaction.
  • A further participant draws a parallel to the reaction between NaOH and HCl, questioning how to determine which ions react, specifically H+ and OH- versus Na+ and Cl-.
  • One response explains that the full ionization of NaOH and HCl leads to the conclusion that H+ and OH- are the reacting species, while Na+ and Cl- are spectator ions.
  • Another participant emphasizes that knowing which ions react is fundamental to chemistry and mentions the common neutralization reaction between H+ and OH-.
  • There is a reiteration of the equilibrium considerations for the reactions involved, highlighting the complexity of the situation.

Areas of Agreement / Disagreement

Participants express varying views on the roles of different ions in the reactions discussed. While some agree on the concept of spectator ions, there is no consensus on the broader implications of solubility rules and their application to other reactions.

Contextual Notes

Participants mention solubility rules and equilibrium constants but do not provide specific details or conclusions regarding their implications for the reactions discussed.

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Homework Statement


An ionic equation is an equation that shows the chemical species that actually take part in a reaction.
Ex. Pb(NO3)2 (aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
So the ionic equation for this reaction is
Pb2+ (aq) + 2I- (aq) -> PbI2 (s)
How do we know that Pb2+ and I- takes part in the reaction? What about K+ and NO3- ?

Homework Equations

The Attempt at a Solution

 
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KNO3 is pretty soluble. So is KI. Perhaps Pb(NO3)2 as well. So except the ions that precipitate, the ions in solution don't even notice what's "happening".
 
BvU said:
KNO3 is pretty soluble. So is KI. Perhaps Pb(NO3)2 as well. So except the ions that precipitate, the ions in solution don't even notice what's "happening".
Ok...then what about NaOH (aq) + HCl (aq) -> NaCl (aq) + H2O (l)
How do we know that H+ and OH- are the ions that react? And not Na+ and Cl-?
 
We know because -- if everything happens in solution, so full ionization (*) and no solids --
the NaOH (aq) on the left stands for Na+ plus OH-
The HCl (aq) on the left for H+ plus Cl-
NaCl (aq) on the right stands for Na+ plus Cl-
And if we leave out the common things on left and right all that happens is H+ + OH- → H2O

So much for the simplistic answer. Reality is a lot more complex, but this case can be untangled by considering it as a set of equilibria:

NaOH (aq) ↔ Na+ + OH-
HCl (aq) ↔ H+ + Cl-
H2O ↔ H+ + OH-
​
with an equilibrium constant for each of these.
And the k for water is so small compared to the other two that it makes the last equilibrium reaction "shift" to the left.
 
Knowing what reacts and when is what chemistry is about. Some things you just have to remember - H+ + OH- is a neutralization reaction, quite common.

What do the solubility rules say about NaCl solubility? Do you expect it to precipitate?
 
BvU said:
We know because -- if everything happens in solution, so full ionization (*) and no solids --
the NaOH (aq) on the left stands for Na+ plus OH-
The HCl (aq) on the left for H+ plus Cl-
NaCl (aq) on the right stands for Na+ plus Cl-
And if we leave out the common things on left and right all that happens is H+ + OH- → H2O

So much for the simplistic answer. Reality is a lot more complex, but this case can be untangled by considering it as a set of equilibria:

NaOH (aq) ↔ Na+ + OH-
HCl (aq) ↔ H+ + Cl-
H2O ↔ H+ + OH-
​
with an equilibrium constant for each of these.
And the k for water is so small compared to the other two that it makes the last equilibrium reaction "shift" to the left.
Oh I see. Thanks!
 
Borek said:
Knowing what reacts and when is what chemistry is about. Some things you just have to remember - H+ + OH- is a neutralization reaction, quite common.

What do the solubility rules say about NaCl solubility? Do you expect it to precipitate?
Nope.. Ok I think I've got it thanks
 

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