Problem 12.69

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Andy Liao 1B
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Problem 12.69

Postby Andy Liao 1B » Mon Dec 04, 2017 7:57 pm

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

Postby Chem_Mod » Tue Dec 05, 2017 12:25 am

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
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Re: Problem 12.69

Postby 304922790 » Tue Dec 05, 2017 10:14 pm

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
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Re: Problem 12.69

Postby 304922790 » Tue Dec 05, 2017 10:34 pm

Sorry never mind, I found the value on table 12.8

Magdalena Palavecino 1A
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Re: Problem 12.69

Postby Magdalena Palavecino 1A » Tue Dec 05, 2017 10:49 pm

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
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Re: Problem 12.69

Postby Rosa Munoz 2E » Sun Jan 19, 2020 5:27 pm

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

AlyshaP_2B
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Re: Problem 12.69

Postby AlyshaP_2B » Sun Jan 19, 2020 7:48 pm

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

Nan_Guan_1L
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Re: Problem 12.69

Postby Nan_Guan_1L » Thu Jan 28, 2021 9:21 pm

to be exact, NH3 and NH4+ are a conjugate acid and base pair, but you can use Kb of NH3 to calculate Ka of NH4Cl because Cl- would mainly be inert in water.

Zoe Kaiser
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Re: Problem 12.69

Postby Zoe Kaiser » Sat Jan 27, 2024 10:21 pm

Wait how are we supposed to get the Ka value for part b? I cannot find it on the table that they told us to use and I can't find the table that someone mentioned.


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