Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: stefufufu on October 30, 2013, 08:42:08 AM
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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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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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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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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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Ka = x.x/(c-x) = x2/c
This doesn't seem right. Using appropriate formatting will help.
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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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Looks OK, I got 2.38.
You have used approximate method - you should check if the approximation is allowed.
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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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Beware - it won't work for 0.01 or 0.001 M solutions of formic acid!
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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. :)