The pH of blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` concentrations. What volume of 5 M `NaHCO_(3)` solution, shnould be mixed with 10 mL sample of blood, which is 2 M in `H_(2)CO_(3)` in order to maintain a pH of `7.4 (K_(a)for H_(2)CO_(3)`in blood =`7.8xx10^(-7)) `

Explanation

`pH = pK_(a) + log. (["Salt"])/(["Acid"])`
`7.4= - log. (7.8xx10^(-7)) + log. ([NaHCO_(3)])/([H_(2)CO_(3)])`
`log .([NaHCO_(3)])/([H_(2)CO_(3)])=7.4+ (-7+0.8921)=1.2921 or ([NaHCO_(3)])/([H_(2)CO_(3)]) = ` Antilog 1.2921 = 19.59
`H_(2)CO_(3)` present in 10 ml of blood `(2 M H_(2)CO_(3))=(2)/(1000)xx10=0.02` mole
`:. NaHCO_(3)` that should be present `=19.59xx0.02 = 0.3918 ` mole
Volume of 5 M `NaHCO_(3)` required `= (1000)/(5) xx 0.3918 = 78.36` ml