The 1st dissociation of carbonic acid is given by:
H_2CO_3rightleftharpoonsHCO_3^(-)+H^+H2CO3⇌HCO−3+H+
For which:
K_((app))=([HCO_3^-][H^+])/([H_2CO_3])K(app)=[HCO−3][H+][H2CO3]
K_((app))K(app) is the value which takes into account the dissolved carbon dioxide. It is equal to 4.0xx10^(-7)"mol/l"4.0×10−7mol/l at 25^@"C"25∘C.
The value given in the question looks incorrect. It seems very big.
I will not use the 2nd dissociation.
Rearranging gives:
[H^+]=K_((app))xx[[H_2CO_3]]/[[HCO_3^(-)]][H+]=K(app)×[H2CO3][HCO−3]
:.[[H_2CO_3]]/[[HCO_3^(-)]]=[H^+]/K_((app))
pH=7.4:.[H^+]=4.0xx10^(-8)"mol/l"
:.[[H_2CO_3]]/[[HCO_3^-]]=(4.0xx10^(-8))/(4.0xx10^(-7))=0.1
Since the total volume of the buffer is common to both we can write:
n_(H_2CO_3)/n_(HCO_3^(-))=0.1
Where n refers to the number of moles.
We know that c=n/v so the number of moles is given by:
n=cxxv
So we can write:
(2xx10/1000)/(5_(V_(HCO_3^-)))=0.1
:.5V_(HCO_3^-)=0.02/0.1
V_(HCO_3^-)=(0.02)/(5xx0.1)=0.04"L"
=40"ml"
