6th edition, 12.69

Moderators: Chem_Mod, Chem_Admin

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

6th edition, 12.69

Postby Tyra Nguyen 4H » Wed Jan 23, 2019 5:25 pm

Not sure how to approach this problem, as the Kb values aren't given in any tables?:

Calculate the pH of each of the following solutions: (a) 0.19 m NH4Cl(aq); (b) 0.055 m AlCl3(aq).

Henry_Phan_4L
Posts: 68
Joined: Fri Sep 28, 2018 12:24 am

Re: 6th edition, 12.69

Postby Henry_Phan_4L » Wed Jan 23, 2019 5:39 pm

So the first step is to take off a Cl from the compound, because it's an ionic compound and ionic compounds fully dissociate. So for NH4Cl, it would turn into NH4+.
Then do the chemical equation if you added H2O to the ion. The products should be NH3 and H3O+.
From there you can do an ICE table.

Oh to get Ka you can divide Kw by Kb. But yeah try to find the Kb for NH4+.

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

Re: 6th edition, 12.69

Postby Tyra Nguyen 4H » Wed Jan 23, 2019 8:08 pm

Henry_Phan_4L wrote:So the first step is to take off a Cl from the compound, because it's an ionic compound and ionic compounds fully dissociate. So for NH4Cl, it would turn into NH4+.
Then do the chemical equation if you added H2O to the ion. The products should be NH3 and H3O+.
From there you can do an ICE table.

Oh to get Ka you can divide Kw by Kb. But yeah try to find the Kb for NH4+.


There is only a Kb for NH3, so I still do not know what to do.

dgerges 4H
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

Re: 6th edition, 12.69

Postby dgerges 4H » Wed Jan 23, 2019 8:20 pm

^^divide kw by the kb for nh3 and you'll get the ka for nh4+


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest