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Topic: pKa value, I'm stuck in the task  (Read 1543 times)

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Offline payata

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pKa value, I'm stuck in the task
« on: May 03, 2019, 08:54:04 AM »
 ???

Hey Guys,

I just started training with exercises, pH and pKa value
I wanted to ask someone to help me understand, and to explain step by step how they got value of : 6.3 x 10-5

Here is a task:
The acid dissociation constant (Ka) for benzoic acid is 6.3 x 10-5. Find the pH of a 0.35 M solution of benzoic acid.

pKa = [H+][ C7H3O2-]/[HC7H3O2] = (x)(x)/0,35 = 6.3 x 10-5

Thank you in advance!



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Re: pKa value, I'm stuck in the task
« Reply #1 on: May 03, 2019, 09:51:14 AM »
Not sure what you are asking, Ka is given, doesn't matter where they got it from.

Plus,

pKa = [H+][ C7H3O2-]/[HC7H3O2] = (x)(x)/0,35 = 6.3 x 10-5

This is wrong in two ways. First, it is not pKa, it is Ka on the left. Second - going to the place where x appears requires assuming x << 0.35, so the equal sign should be replaced by ≈ (approximately equal).
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Re: pKa value, I'm stuck in the task
« Reply #2 on: May 03, 2019, 10:05:32 AM »
The solubility of benzoic acid at 25°C (298 K) is very small. The saturated solution at this temperature is below 0.03 M
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Offline payata

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Re: pKa value, I'm stuck in the task
« Reply #3 on: May 05, 2019, 02:56:24 PM »
Oh, ok . I am a bit still confuse:

for example, teacher gave us to solve, find pH of 0.1M acetic acid.That's it, no more info. Therefore I was looking for pKs value which is 4, 75 . but no:

When I look to internet, and everywhere is given Ka value?, How can I know that I need to look up for ka value? how to know the diffrents where to you pka, ka. He is also using aberr. of pKs, is it the same as pKa and pKb  :-[

If somebody somebody can explain it very easy I would be grateful  ???

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Re: pKa value, I'm stuck in the task
« Reply #4 on: May 05, 2019, 03:12:47 PM »
pKa=-log(Ka)

Ka=10-pKa
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Offline payata

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Re: pKa value, I'm stuck in the task
« Reply #5 on: May 05, 2019, 04:05:19 PM »
So, to find Ka, I will use Ka=10-pKa, and to find to find pKa: pKa=-log(Ka) .. ?

And what about pKb, pKs?

In last example, why I have to use Ka value?

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Re: pKa value, I'm stuck in the task
« Reply #6 on: May 05, 2019, 04:44:18 PM »
So, to find Ka, I will use Ka=10-pKa, and to find to find pKa: pKa=-log(Ka) .. ?

By definition.

Quote
And what about pKb, pKs?

pKwhatever always means -log(Kwhatever).

Quote
In last example, why I have to use Ka value?

Because that's what directly describes the dissociation equilibrium.

« Last Edit: May 06, 2019, 03:18:03 AM by Borek »
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Offline payata

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Re: pKa value, I'm stuck in the task
« Reply #7 on: May 05, 2019, 04:45:26 PM »
Clear! Thank you!  :)

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