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Author Topic: pH from pKa value  (Read 6089 times)

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stefufufu

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pH from pKa value
« on: October 30, 2013, 02:42:08 AM »

What is the pH of:
A) 0.10 M formic acid pKa = 3.75
B) 0.05 M dichloroethanoic acid pKa = 1.48
C) 0.015 M carbonic acid pKa1 = 6.37 [6]


a) Ka = -log pKa = 10^-3.75 = 1.78x10⁻⁴ = (1x10⁻¹⁴)/(1.78x10⁻⁴) = 5.61x10⁻¹¹
   √(0.1 x 5.61x10⁻¹¹) = 2.37x10⁻⁶
   
pOH = -log(2.37x10⁻⁶) = 5.6
pH = 14 – pOH = 8.4
   
b) Ka = -log pKa = 10^-1.48 = 3.31x10⁻² = (1x10⁻¹⁴)/(3.31x10⁻²) = 3.02x10⁻¹³
   √(0.05 x 3.02x10⁻¹³) = 1.23x10⁻⁷

pOH = -log(1.23x10⁻⁷) = 6.9
pH = 14 – pOH = 7.1

c) Ka = -log pKa = 10^-6.37 = 4.27x10⁻⁷ = (1x10⁻¹⁴)/(4.27x10⁻⁷) = 2.34x10⁻⁸
   √(0.015 x 2.34x10⁻⁸) = 1.87x10⁻⁵

pOH = -log(1.87x10⁻⁵) = 4.7
pH = 14 – pOH = 9.3

I used notes and the internet for the equations, which seemed to work fine until I googled the pH of the acids and it makes my answers completely wrong.

I originally thought the pH could be worked out by doing the -log of Ka, but then M concentration was involved so not really sure which way to go.
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Borek

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Re: pH from pKa value
« Reply #1 on: October 30, 2013, 03:03:49 AM »

Hard to say what you are doing, you just listed some numbers without any explanation. Please start again with just a first problem (0.1M formic acid) and explain what you are doing and why.

Use [sup][/sup] tags to format exponents. 10[sup]-4[/sup] is rendered as 10-4.

I originally thought the pH could be worked out by doing the -log of Ka, but then M concentration was involved so not really sure which way to go.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
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stefufufu

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Re: pH from pKa value
« Reply #2 on: October 30, 2013, 03:16:26 AM »

Okay,
A) 0.10 M formic acid pKa = 3.75

I used the pKa value and did the -log to get the Ka value.
Ka = -log pKa = 10^-3.75 = 1.78x10-4

I have just realised here, I have used 1x1014, which is for working with bases?? Without that...

OH- = √(M x Ka) = √(0.10 x 1.78x10-4) = 4.22x10-3

pOH = -log(OH-) = -log(4.22x10-3) = 2.4
pH = 14 - pOH = 11.6.

Oh dear, I think I have just confused myself. I will have a look at that wesbite.
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stefufufu

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Re: pH from pKa value
« Reply #3 on: October 30, 2013, 03:44:05 AM »

Okay, let me try again.
Formic acid, pKa = 3.75 c = 0.10M.

Ka = 10-pka = 10-3.75 = 1.78x10-4.

Ka = x.x/(c-x) = x2/c
x = √(Ka x C) =  √(1.78-4 x 0.1) = 4.22x10-3.

0.1 - 4.22x10-3 = 9.58x10-2 ~ 0.1.

pH = -log(9.58x10-2) = 1.01.

?
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Corribus

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Re: pH from pKa value
« Reply #4 on: October 30, 2013, 03:51:00 AM »

Ka = x.x/(c-x) = x2/c
This doesn't seem right.  Using appropriate formatting will help.
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stefufufu

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Re: pH from pKa value
« Reply #5 on: October 30, 2013, 04:00:03 AM »

Ka = 10-pka = 10-3.75 = 1.78x10-4.

Ka = x.x/(c-x), where x = [H+]

Ka ~ x2/c

x = √(Ka.C) =  √(1.78x10-4 x 0.1) = 4.22x10-3.

0.1 - 4.22x10-3 = 9.58x10-2 ~ 0.1.

[H+] = x = 4.22x10-3

pH = -log [H+] = -log(4.22x10-3) = 2.37

?
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Borek

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Re: pH from pKa value
« Reply #6 on: October 30, 2013, 04:13:28 AM »

Looks OK, I got 2.38.

You have used approximate method - you should check if the approximation is allowed.
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stefufufu

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Re: pH from pKa value
« Reply #7 on: October 30, 2013, 04:15:20 AM »

Well, finally getting there at least.

This method was partly in my lecturers slides, so I am assuming it is okay!
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Borek

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Re: pH from pKa value
« Reply #8 on: October 30, 2013, 04:37:23 AM »

Beware - it won't work for 0.01 or 0.001 M solutions of formic acid!
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Corribus

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Re: pH from pKa value
« Reply #9 on: October 30, 2013, 05:52:39 AM »

Ka = x.x/(c-x), where x = [H+]

Ka ~ x2/c
Like Borek said, this approximation will not work at all times and it's not really that necessary. Easy to solve quadratic equation to get real result. Up to you, of course. :)
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