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"#
