collisoo
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Calculating pH
Given that the pKa of acetic acid is 4.7, what are the pH values of the three buffers? Satisfy yourself from your answer that these solutions have
increasing power of elution from DEAE-cellulose.
Buffer I; 20mM sodium acetate + 5mM acetic acid => 5E-3 M acetic acid
Buffer II; 40mM sodium acetate + 40mM acetic acid => 40E-3 M acetic acid
Buffer III; 100mM sodium acetate + 100mM acetic acid => 100E-3 M acetic acid
Ka of acetic acid => pKa = -log[Ka] = > 1.995E-5
x = √( (Ka)(no.of moles)
pH = -log[x]
Buffer I; x = 3.158322339E-4, pH = 3.50
Buffer II; x = 8.933084574E-4, pH = 3.05
Buffer III; x = 1.412444689E-3, pH = 2.85
I don't think this is right, but I can't see where I'm going wrong. Help, please, anyone?
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unionised
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This might help.
http://www.sparknotes.com/chemistry/acidsbases/buffers/secti...
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collisoo
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Well duh
Buffer I; pH = 4.7 + log([20E-3]/[5E-3]) = 5.3
Buffer II; pH = 4.7 + log ([40E-3]/[40E-3]) = 4.7
Buffer III; pH = 4.7 + log ([100E-3]/[100E-3]) = 4.7
Thanks for that. I knew I should have stuck with the Henderson-Hasselbalch equation
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Magpie
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A weak organic acid has a pKa of 4.2. It's dissociation in water can be represented as:
HA = A- + H+
The Henderson-Hasselbalch equation is as follows:
pH = pKa + log{[A-]/[HA]}
To get ~100% of the acid into the non-ionized form the pH must be ≤ 2.2, ie
pH = 4.2 + log (1/100) = 4.2 - 2 = 2.2
This acid is in solution at pH = 4.
Is it correct to say that the pH of this solution must be brought down to ≤ 2.2 to get near total precipitation?
The single most important condition for a successful synthesis is good mixing - Nicodem
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