## 7th edition: 6D.15

Aurbal Popal
Posts: 63
Joined: Fri Sep 28, 2018 12:27 am

### 7th edition: 6D.15

Calculate the pH of:
a) .19 NH4Cl
b) .055 M AlCl3

I saw a couple of posts that addressed this question, especially part b. I understand that Cl is not reactive and that water dissolves the aluminum, but that does not happen for NH4Cl so I was kind of confused as to why they were solved differently. Is it because aluminum is usually a solid/metal? Does the fact that it is aqueous not mean anything? How are we supposed to know whether we can ignore the Cl or when we have to show that water dissolves it? Also, how do you know how many waters bond with the aluminum?

Ethan Breaux 2F
Posts: 63
Joined: Sat Sep 29, 2018 12:16 am

### Re: 7th edition: 6D.15

Pretty sure (aq) just means its like surround by water or in water. The problems are also solved basically the same way and there is a chart with values and like info on Al^3+ and how it equals Al(H20)6^3+ (I think that that would be given to us)

Jake Gordon 1A
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

### Re: 7th edition: 6D.15

I agree with the person above, also Cl by itself doesn't affect pH. The only time it could be part of a dissociation that changes pH would be something like HClO, but even this is because of the hydrogen not the chlorine. For NH4Cl we are focused on NH4 + dissociating into H3O+ and NH3 the Cl doesn't make OH- or H + ions so it doesn't affect pH

I hope this is what you are asking.