Question

Calculate the pH of a 0.0174 M solution of arginine hydrochloride (arginine·HCl, H2Arg ). Arginine has pKa values of 1.823 (pKa1), 8.991 (pKa2), and 12.01 (pKa3).Calculate the concentration of each species of arginine in the solution.

Answer 1

Arginine•HCl or H2Arg partially dissociates in solution as per the reaction,

H2Arg +H2O↔HArg +H3O+    ( pka3=12.01)   [side chain   -NH3⁺ group loses proton]

HArg +H2O↔HArg +H3O+ (pka2=8.991)   [alpha-NH3⁺ group loses proton]

Arg+H2O↔Arg- +H3O+     (pka1=1.823) [-COOH grp looses proton]

To calculate pH prepare ICE table

	[H2Arg]	[H3O+]	[HArg]
Initial	0.0174 M	0	0
change	-x	+x	+x
equilibrium	0.0174-x	x	x

Ka3=[H3O+][HArg]/[H2Arg]

pKa3=-logka3=12.01

ka3=10^-12.01=9.772*10^-13

9.772*10^-13=x^2/(0.0174-x) [ignore x<<<0.0174 as dissociation is very small]

Or, 9.772*10^-13=x^2/(0.0174)

X^2=0.17*10^-13=1.7*10^-14

X=sqrt(1.7*10^-14)=1.30*10^-7 M=[H3O+]=[HArg]

[HArg] =1.30*10^-7 M

[H2Arg]=0.0174-x=0.0174M (as x is very small comparatively)

Next HArg dissociates,

	[HArg]	[H3O+]	[Arg]
Initial	1.3*10^-13	1.3*10^-13	0
change	-x	+x	+x
equilibrium	1.3*10^-13-x	1.3*10^-13+x	x

Pka2=8.991

Ka2=10^-8.991=1.02*10^-9=[H3O+][Arg]/[HArg]

1.02*10^-9=x*(1.3*10^-13+x)/ (1.3*10^-13-x)

Or, 1.02*10^-9=x*(1.3*10^-13)/ (1.3*10^-13) [ x<<than 1.3*10^-13]

Or,x=1.02*10^-9 M=ka=[H3O+]=[Arg]

[Arg]= 1.02*10^-9 M

	[Arg]	[H3O+]	[Arg-]
Initial	1.02*10^-9	1.02*10^-9	0
change	-x	+x	+x
equilibrium	1.02*10^-9 -x	1.02*10^-9 +x	x

Ka1=10^-ka1=10^-1.823=0.015

Ka1=[H3O+][Arg-]/[Arg]

Or, 0.015=x*(1.02*10^-9 +x)/( 1.02*10^-9 –x)

Or, 0.015=x*(1.02*10^-9 )/( 1.02*10^-9 )

Or,0.015=X=[H3O+]=[Arg-]

[Arg-]=0.015

pH=-log[H3O+]=-log (0.015)=1.83

pH=1.83