Answer:
The major ionization state is at pH of 6,2.
Explanation:
Succinic acid HOOC(CH₂)₂COOH has as equilibrium:
HOOC(CH₂)₂COOH ⇄ HOOC(CH₂)₂COO⁻ + H⁺; pka = 4,21
The Henderson-Hasselbalch formula:
pH = pka + log₁₀ [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
For a pH of 3,2:
3,2 = 4,21 + log₁₀ [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
0,0977 = [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
For a pH of 4,2:
4,2 = 4,21 + log₁₀ [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
0,977 = [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
For a pH of 5,2:
5,2 = 4,21 + log₁₀ [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
9,77 = [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
For a pH of 6,2:
6,2 = 4,21 + log₁₀ [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
97,7 = [tex]\frac{[HOOC(CH_{2})_{2}COO^-]}{[HOOC(CH_{2})_{2}COOH]}[/tex]
Thus, the major ionization state is at pH of 6,2.
