The 1st dissociation of carbonic acid is given by:

#H_2CO_3rightleftharpoonsHCO_3^(-)+H^+#

For which:

#K_((app))=([HCO_3^-][H^+])/([H_2CO_3])#

#K_((app))# is the value which takes into account the dissolved carbon dioxide. It is equal to #4.0xx10^(-7)"mol/l"# at #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_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"#