Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tjpj77 on September 05, 2012, 03:26:02 AM
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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!
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[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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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....
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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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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!!!!!!!!!!!!!