K values with acid and base reactions

[MathewsMD]

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?

 

[Chestermiller]

[h^+][oh^-]=10^{-14}

 

[MathewsMD]

[h^+][oh^-]=10^{-14}

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?

 

[Chestermiller]

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

 

[MathewsMD]

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 K_b by K_w to get the second reaction I specified?

 

[Borek]

K_a \times K_b = K_w

http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory

Ah, the joys of reinventing the wheel