How to compute pH of ((COOH)2) Oxalic acid?
Posted: Wed Apr 21, 2021 6:55 am
pH of oxalic acid is given 5.4e-2 in table of equilibrium constants for acids and bases in aqueous solution. But, on other chemistry website , it is stated as
'' The constant acid dissociation of oxalic acid is 5.60 between 10⁻² and 5.42 between 10⁻⁵. As oxalic acid is a polyprotic acid, there are two values and it has a chemical formula HOOC-COOH."
What does this mean?
HO2CCO2H⇌H⁺ + HOOCCOO⁻ 📖 Assuming 1.0 M of ((COOH)2) aqueous solution, Ka=54e-2= x²/(1-x)→ x=0.207
So,pH= -log(0.207) =0684
But in table, pH=1.27 How is that?
'' The constant acid dissociation of oxalic acid is 5.60 between 10⁻² and 5.42 between 10⁻⁵. As oxalic acid is a polyprotic acid, there are two values and it has a chemical formula HOOC-COOH."
What does this mean?
HO2CCO2H⇌H⁺ + HOOCCOO⁻ 📖 Assuming 1.0 M of ((COOH)2) aqueous solution, Ka=54e-2= x²/(1-x)→ x=0.207
So,pH= -log(0.207) =0684
But in table, pH=1.27 How is that?