using kA vs kB when solving for the pH of a salt solution

[LyahFitzpatrick 2E]

In the example problem in Friday's lecture for finding the pH of a NH4Cl solution, NH4Cl acted as an acid (as a proton donor) so we solved for Ka by using the equation Ka*(Kb for NH3)=Kw. However, since we set up the equation with NH4 in the reactants and NH3 in the product, wouldn't we need to use the inverse of Kb to solve for Ka? or does the Ka*Kb=Kw equation already take that relationship into account?

[Kayley Steele 3D]

The equation KaxKb= water can be used to convert Ka to Kb or vise versa no matter which way the reaction is going so you don't need to take the inverse!

[Alex Luong 3H]

You could use the equation kw = (ka)(kb) to switch in between ka and kb. So if you were given ka, but need kb, you could use that equation to switch in between the 2.