[tjpj77]

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!

[Borek]

[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.

[tjpj77]

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....

[Borek]

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.

[tjpj77]

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!!!!!!!!!!!!!