Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4�3H2O, MW = 248.32 g/mol) must be added to 530.0 mL of a 0.0449 M succinic acid solution to produce a pH of 5.835? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Answer 1

succinic acid is a diprotic acid

Concentration of succinic acid = 0.0449 M

pKa1 = 4.207

Ka1 = 10^-4.207 = 6.208x10^-5

pKa2 = 5.636

Ka2 = 10^-5.636 = 2.31x10^-6

Let the succinic acid dissociates as-

[H2A]      H⁺   + HA^-

Initial 0.0449 0 0

Final 0.0449 - x x x

Ka1 = x*x / (0.0449 - x)

6.208x10^-5 = x² /  (0.0449 - x)

2.787x10^-6 - 6.208x10^-5 *x = x²

x² + 6.208x10^-5 *x - 2.787x10^-6 = 0

x = 1.638x10^-3 M

[H+] = [HA-] = 1.638x10^-3 M

Again,

HA- H+ + A2-

initial 1.638x10^-3 0 0

final 1.638x10^-3 - x x x

Ka2 = x² / (1.638x10^-3 - x)

2.31x10^-6 = x² / (1.638x10^-3 - x)

3.78x10^-9 - 2.31x10^-6 *x = x²

x² + 2.31x10^-6 *x - 3.78x10^-9 = 0

x = 6.03x10^-5 M

[H+] = [A2-] = 6.03x10^-5 M

Total H+ ions concentration = 1.638x10^-3 M + 6.03x10^-5 M

= 1.698x10^-3 M

volume of succinic acid = 530 mL = 0.53 L

moles of H+ ions = 0.53*1.698x10^-3

= 8.99x10^-4

moles of dipotassium succinate trihydrate required = 8.99x10^-4

mass of dipotassium succinate trihydrate required = 248.32 *  8.99x10^-4

= 0.2234 grams