Calculating the pH of a salt solution (lecture example)

[905574994]

Hi guys,

We did this example during lecture:
What is the ph of .15 M NH4Cl? KB for NH3=1.8 x 10^5

For this question, how did we know that we needed to use the kA value rather than the kB value?

Thanks

[KPINTO 1B]

Hey! We know we need to use the Ka rather than the Kb, because when NH4Cl dissociates, it turns into NH4+ and Cl-. We can forget about Cl- in this example because it is a spectator ion, so then we have NH4+ and that will react with H20, to form H3O+ and NH3. Thus, because we are making H30+, we need to use the Ka rather than the Kb. If NH4+ was our product and we add OH- as the other product, then you would use Kb. Hope this helps !

[KyleNagasawaDisc3C_Chem 14B2022W_]

Hey,

Ka and Kb can be relatively ambiguous as to what species they refer to, and I understand your confusion. Like if we receive Kb (NH3), it would be easy to confuse this as to refer to NH3 or NH2- or even ka(NH3) to refer to NH3 or NH4+. To best evaluate which value it refers to, you can assess the magnitude of the constant. More stables species will be preferred.

In answering your question, we know that NH4+ (dissociation of the salt generated NH4+ and Cl-) will function as an acid much more readily than a base (wants to remove additional charge), generating NH3 and H+. Kb refers to an equilibrium constant of a base, while ka refers to the equilbrium constant of an acid deprotonating. To calculate kb from ka, see the autoionization of water formula.