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Topic: Produce this solution {ph buffer succinate}  (Read 6186 times)

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Offline tjpj77

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Produce this solution {ph buffer succinate}
« on: September 05, 2012, 03:26:02 AM »
How many grams of sodium succinate (140 amu) and disodium succinate (162 amu) must be added to 1L of water to produce a solution with pH 6.0 and a total solute concentration of 50mM?

I use pH = pKa + log ([A-]/[HA]) and since pKa of succinic acid averages to 4.8, I get 6 = 4.8 + log ([A-]/[HA])
and so 1.2  = log ([A-]/[HA]).

but then I get totally stuck...

please *delete me*!!

thank you!
« Last Edit: September 05, 2012, 06:59:29 AM by Arkcon »

Offline Borek

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Re: Produce this solution...
« Reply #1 on: September 05, 2012, 03:42:58 AM »
[A-]+[HA]=50 mM

But you can't average pKas, you have to use one of these values - the one that is closer to the required pH.
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Offline tjpj77

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Re: Produce this solution...
« Reply #2 on: September 05, 2012, 03:53:47 AM »
got it re. pKa value to use...
so, 6.0 = 4.2 + (50mM -HA)/HA?
For sure, 1.8 = [A-/HA]
Then I get HA = 18 and A = 32, assuming my second line above is right, which it's not. i'm totally stuck on A- + HA = 50mM...I don't know what to do with that....

Offline Borek

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Re: Produce this solution...
« Reply #3 on: September 05, 2012, 04:21:41 AM »
6.0 = 4.2 + (50mM -HA)/HA?

So 4.2 is closer to 6 than 5.6?

A- + HA = 50mM...I don't know what to do with that...

This time you already used this information, 50mM-HA is an obvious conclusion of this equation.
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Offline tjpj77

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Re: Produce this solution...
« Reply #4 on: September 05, 2012, 04:31:19 AM »
very cool...
so, 0.4HA = 50mM
HA = 35.7 mM = sodium succinate
A = 14.3 mM = disodium succinate
after the math, i need 5 g of sodium succinate and 2.3 g of disodium succinate!
very cool...
thank you!!!!!!!!!!!!!

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