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### Topic: % dissociation of a base at different temperatures  (Read 3034 times)

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#### Durlag

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##### % dissociation of a base at different temperatures
« on: March 21, 2014, 06:11:53 AM »
Just checking something:

A base has pKa of 14 at 25°C and 10,4 at 40°C.

We also know that pH of water equals 7 at 25°C and 6,77 at 40°C (source:http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product).

According to the equation [acid]/[base] = 10pKa-pH (derived from the Henderson-Hasselbalch equation), does this mean that:
at 25°C, pKa - pH = 7 so % dissociation = 9,99.10-6 %
at 40°C, pKa - pH = 3.63 so % dissociation = 0.023 %

?

Thanks!

#### Borek

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##### Re: % dissociation of a base at different temperatures
« Reply #1 on: March 21, 2014, 01:04:21 PM »
If you put a base into water solution its pH is no longer that of a pure water.
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#### Durlag

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##### Re: % dissociation of a base at different temperatures
« Reply #2 on: March 21, 2014, 02:56:40 PM »
Yes, but don't you calculate the new pH after having calculated the % dissociation of the base that will happen when you will add that base in water?

#### Borek

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##### Re: % dissociation of a base at different temperatures
« Reply #3 on: March 21, 2014, 04:05:40 PM »
Technically it is the same calculation. Whether you start with finding pH to find % dissociation, or whether you find % dissociation to find pH, you are solving exactly the same problem - you find the dissociation equilibrium.
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#### Durlag

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##### Re: % dissociation of a base at different temperatures
« Reply #4 on: March 23, 2014, 10:59:47 AM »
I get it.

First case:
-You calculate the % dissociation as I have done in the original post.
-Then, for a base, % dissociation = x/[B ]0*100 with x=[BH+]=[OH-]. If you know [B ]0 you can calculate [OH-] the pOH and then pH.

Second case:
- Knowing [B ]0, you do an ICE table, and solve for x=[BH+]=[OH-].
-Then you calculate % dissociation according to formula.

However, my question is still valid, at different temperatures, will the % dissociation be different?
« Last Edit: March 23, 2014, 01:16:05 PM by Durlag »

#### Borek

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##### Re: % dissociation of a base at different temperatures
« Reply #5 on: March 23, 2014, 01:44:50 PM »
Dissociation constant changes, so yes, the dissociation % changes as well.
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#### Durlag

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##### Re: % dissociation of a base at different temperatures
« Reply #6 on: March 24, 2014, 03:06:12 AM »
Nice! I still hadn't made the connection bewteen Ka/Kb and dissociation %. I wasn't seeing their true meaning (dissociation constants). It's funny how sometimes you get an idea out of words that is different than the original idea which the words were used to describe. A light bulb just went on in my head! Haha. Thanks for your he|p!

However, at 40°C, should you take pH 6,77 as the pH of water? I would say yes, since it reflects [H+] and thus [OH-], but I'd like confirmation.
« Last Edit: March 24, 2014, 03:21:31 AM by Durlag »

#### Borek

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##### Re: % dissociation of a base at different temperatures
« Reply #7 on: March 24, 2014, 03:57:47 AM »
However, at 40°C, should you take pH 6,77 as the pH of water? I would say yes, since it reflects [H+] and thus [OH-], but I'd like confirmation.

I already told you pH of PURE water is not related to the problem.
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#### Durlag

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##### Re: % dissociation of a base at different temperatures
« Reply #8 on: March 24, 2014, 10:19:37 AM »
Ok, I think I get it. The pH in the HH equation is the pH at equilibrium. This means you can't calculate a % dissociation from knowing only the pKa and the pH of pure water. You have to find the pH at equilibrium thanks to an ICE table first. This means you need to know [B ]0.

However, this also means the HH equation is useless in this case, since you can calculate the % dissociation without using the HH equation.

This also means my calculations in my original post are wrong.

I'm not 100% sure about all this. I'm wrong a lot of the time.
« Last Edit: March 24, 2014, 11:41:28 AM by Durlag »