Answer:
pH = 5,407
Explanation:
Arginine is an amphoteric solution. The equilibriums of H₂Arg⁺ are:
H₃Arg²⁺ ⇄ H₂Arg⁺ + H⁺ pka1 = 1,823; k1 = [tex]10^{-1,823}[/tex] = 0,01503
k1 = [H₂Arg⁺] [H⁺] / [H₃Arg²⁺] (1)
H₂Arg⁺ ⇄ HArg + H⁺ pka2 = 8,991; k2 = [tex]10^{-8,991}[/tex] = 1,021x10⁻⁹
k2 = [HArg] [H⁺] / [H₂Arg⁺] (2)
As H₃Arg²⁺ and HArg are produced both from H₂Arg⁺:
2 H₂Arg⁺ ⇄ HArg + H₃Arg²⁺
[HArg] = [H₃Arg²⁺] (3)
Using (3) and replacing (2) in (1) you will obtain:
[H⁺]² = k1×k2
[H⁺]² = [tex] 0,01503* 1,021x10⁻⁹[/tex]
[H⁺] = 3,96x10⁻⁶
As pH = -log[H⁺]
pH = 5,407
I hope it helps!
