Question

The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW = 162 g/mol) must be added to 1 L of water to produce a solution with a pH 5.28 and a total solute concentration of 100 mM? (Assume the total volume remains 1 liter, answer in grams monosodium succinate, grams disodium succinate, respectively.)

Answer 1

Total solute concentration = 100mM = 0.100 M

Let [HA] = sodium succinate and [A^-] = disodium succinate

So, [A^-] + [HA] = 0.100 M => [A^-] = 0.100 M - [HA]

Now, log ( [A^-] / [HA] ) = pH - pK = 5.28 - 5.64

=> [A^-] / [HA] = antilog (-0.36)

=> [A^-] / [HA] =0.44

=> 0.100 M - [HA] / [HA] = 0.44

=> 0.100 M - [HA] = 0.44 [HA]

=> 0.100 M = 0.44 [HA] + [HA]

=> 0.100 M = 1.44 [HA]

=> [HA] =0.100 M / 1.44 M = 0.069 M

[A^-] = 0.100 M - 0.069 M = 0.031 M

Finally,

Grams of sodium succinate = (0.069 mol /L) (140 g/mol) x(1L) = 9.66 g

and Grams of disodium succinate = (0.031 mol/L) (162 g/mol) x(1L) = 5.02 g