[CT]

Hey folks,

We were asked to calculate the pH of these four solutions.


Calculate pH in the following four salt solutions: 

a) CH3COONH4
b) CH3COONa
c) NH4Cl
d) AlCl3

All have 1M consentrations.

Tips: in a) look at the equilibrium reaction: CH3COO- + NH4+ = CH3COOH + NH3

You are given the following acid dissociation constants:

CH3COOH (HAc)    Ka(HAc) = 1.8*10^-5
NH4+    Ka(NH4+) = 5.6*10^-10
AlCl31)    Ka(AlCl3 )= 1.4*10^-5

I really don't understand the procedure. Please explain

Thanks

[Borek]

What is the procedure that you don't understand?

[CT]

I dont understand how you calculate the pH. Also I don´t know how the reaction between each different salt and water looks like.

[Borek]

Sorry, nobody is going to explain everything starting from the very beginning, that's what the books and pH lectures are for. You need to try by yourself, we can help with you details.

[CT]

Okay, I was not aware of the rules of this forum. The topic can be deleted, then.