K values with acid and base reactions

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Discussion Overview

The discussion revolves around the equilibrium constants (K values) associated with acid-base reactions, specifically focusing on the reactions involving ammonia (NH3) and its interactions with water (H2O) and protons (H+). Participants explore the relationships between different K expressions and their implications in calculating concentrations of hydroxide (OH-) and hydrogen ions (H+).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the reactions NH3 + H2O and NH3 + H+ are equivalent and questions whether they share the same Kb value.
  • Another participant states the relationship [H+][OH-] = 10^-14, implying a connection to the ion product of water.
  • There is confusion about the equivalence of the two K expressions, with a participant seeking clarification on why they may not be equivalent despite representing the same reaction.
  • A participant explains that there are two interacting reactions involved and that different equilibrium constants are necessary depending on whether one is calculating OH- or H+ concentrations.
  • One participant inquires if dividing Kb by Kw would yield the second reaction's K value, indicating a potential method for relating the constants.
  • A reference to the relationship Ka × Kb = Kw is made, suggesting a foundational principle in acid-base chemistry.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the equivalence of the K expressions and whether they can be used interchangeably. Multiple competing views remain about how to approach the calculations and the relationships between the constants.

Contextual Notes

Participants note that the second reaction expression may not be in a usable form for direct calculations, and there are discussions about the need to modify K values for different contexts. The discussion also highlights the dependency on the original pH of the solution.

MathewsMD
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Given an aqueous solution (I'm omitting the subscripts):

## NH_3 + H_2O ←→ NH_4^+ + OH^- K_b ##Then we can also say:

## NH_3 + H^+ ←→ NH_4^+ K_b ## as well since they're equivalent reactions, right? They both have the same Kb value as a result, right?

So then, ## K_b = \frac {[NH_4^+][OH^-]}{[NH_3]} = \frac{[NH_4^+]}{[NH_3][H^+]} ##

I would just like to clarify, is this true? It looks like ##[OH^-] = \frac{1}{[H^+]}##

Is this true? Does it take into account the constant "concentration" of the water in this case?
 
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[tex][h^+][oh^-]=10^{-14}[/tex]
 
Chestermiller said:
[tex][h^+][oh^-]=10^{-14}[/tex]

Ok. So are the two K expressions not equivalent? I just don't understand why since they're representing the same reaction, are they not?
 
MathewsMD said:
Ok. So are the two K expressions not equivalent? I just don't understand why since they're representing the same reaction, are they not?
There are actually two interacting reactions involved. One is NH3 + H2O, and the other one is OH+H going to H2O. If you're only interested in finding the OH concentration, and the original solution is pH=7, then you can solve for the new OH concentration using the first equilibrium constant. However, if you also want to find the new H+ concentration, you need to use the second equilibrium constant (OH+H). Your second reaction expression is not in a form that you can use to solve directly and uniquely for the product concentrations (and, of course, the K has to be modified).

Chet
 
Chestermiller said:
There are actually two interacting reactions involved. One is NH3 + H2O, and the other one is OH+H going to H2O. If you're only interested in finding the OH concentration, and the original solution is pH=7, then you can solve for the new OH concentration using the first equilibrium constant. However, if you also want to find the new H+ concentration, you need to use the second equilibrium constant (OH+H). Your second reaction expression is not in a form that you can use to solve directly and uniquely for the product concentrations (and, of course, the K has to be modified).

Chet

Oh okay! Thank you! So would I just divide Kb by Kw to get the second reaction I specified?
 

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