- #1
MathewsMD
- 433
- 7
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?
## 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?