## Problem 12.69

Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Problem 12.69

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

In the solutions manual, the proton transfer equilibria for part (a) is written as NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3. Why isn't it NH4Cl (aq) + H2O (l) ⇌ H3O+ (aq) + NH3Cl- (aq) instead? Is it incorrect to write it like the latter? Also, for part (b) where does the Al(H2O)6 3+ (aq) come from? The proton transfer equilibria for part (b) is Al(H2O)6 3+ (aq) + H2O (l) ⇌ H3O+ (aq) + Al(H2O5)OH 2+ (aq).

Chem_Mod
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### Re: Problem 12.69

For (a), the reason they write it out like that is because Cl- in this case acts as a spectator ion. We tend to cancel out spectator ions when discussing reactions, as they don't contribute anything significant to the reaction. For (b), I believe that's an equivalent way to write out AlCl3 when this species is placed in water (it gets hydrated to form this coordination complex). For this problem, the main idea is to understand how to do the equilibrium problem.

304922790
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am

### Re: Problem 12.69

For part b, the Ka is given as 1.4 x 10^-5. I tried looking for the Ka or Kb (and then converting it to Ka) on the table but I couldn't find it. I was wondering how they got the value.

304922790
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am

### Re: Problem 12.69

Sorry never mind, I found the value on table 12.8

Magdalena Palavecino 1A
Posts: 54
Joined: Fri Sep 29, 2017 7:04 am

### Re: Problem 12.69

How do you go about solving the problem (a) is you only have Kb of NH3 and it is on the products side? Are you supposed to solve for Ka or should you keep Kb and work to the other side?

Rosa Munoz 2E
Posts: 105
Joined: Wed Sep 18, 2019 12:21 am

### Re: Problem 12.69

I am confused as well on why you can use the Kb value for NH3

AlyshaP_2B
Posts: 112
Joined: Wed Sep 18, 2019 12:19 am

### Re: Problem 12.69

Rosa Munoz 2E wrote:I am confused as well on why you can use the Kb value for NH3

You can use the Kb value for NH3 to find the Ka value of NH4Cl because they are a conjugate acid/base pair

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