Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 570.0 mL of a 0.0306 M succinic acid solution to produce a pH of 5.987? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Answer 1

succinic acid (C4H6O4) is a weak acid

The balanced reaction with ICE TABLE

C4H6O4 . .+ . .H2O . .==> . .H3O+ . .+ . C4H5O4-
I......... . . . . . . .0.0306 . . . . . . . . . . . ..... . . . . .0 . . . . . . . . . .0
C......... . . . . . . . .-x . . . . . . . . . . . . ...... . . . . . .x . . . . . . . . . .x
E...... . . . .. . . .0.0306-x . . . . . . . . . ... . . . . . . .x . . . . . . . . . .x

Equilibrium constant expression of the reaction

Ka = 10^-pK1 = 10^-4.207 = 6.21 x 10^-5

Ka = [H3O+][C4H5O4-] / [C4H6O4]

6.21 x 10^-5 = (x)(x) / (0.0306-x)

x^2 = (0.0306-x)(6.21 x 10^-5)
x^2 = 1.90 x 10^-6 - 6.21 x 10^-5x
x^2 + 6.21 x 10^-5x - 1.90 x 10^-6 = 0

x = 0.00135

The given pH (5.979) is greater than pKa2

C4H5O4- and C4H4O4 2- (disuccinate ion).

From Henderson-Hasselbalch equation

pH = pKa2 + log [buffer base] / [buffer acid])

5.987 = 5.636 + log ([C4H4O4 2-] / [C4H5O4-])

0.351 = log ([C4H4O4 2-] / [C4H5O4-])

[C4H4O4 2-] / [C4H5O4-] = 10^0.351 = 2.244

[C2H4O4 2-] = 2.244 x [C4H5O4-]

= 2.244 x 0.00135

= 0.0030294 M

disuccinate salt contains 1 mole of disuccinate for 1 mole of salt

Moles of C4H4O4 2- = moles of K2C4H4O4 x 3H2O

moles of C4H4O4- = molarity x volume

= (0.0030294)(0.570) L

= 0.001726758 mol
moles of K2C4H4O4 * 3H2O = 0.001726758 mol

Mass of K2C4H4O4 * 3H2O = moles x molecular weight

= 0.001726758 mol x 248.32 g/mol

= 0.4287 g