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

Mixing `H_(2)CO_(3)` with `NaHCO_(3)` results in buffer solution.
`pH = pK_(a) + log[NaHCO_(3)]/[H_(2)CO_(3)] = pK_(a)+logn(NaHCO_(3))/n(H_(2)CO_(3))`
`implies 7.4 = -log(7.8 xx 10^(-7)) + log x/20 implies x = 400 m mol`
`implies NaHCO_(3) = 5 xx V implies V = 80 mL`