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

Let the volume of 5 M `NaHCO_3` solution added be x ml.
Number of millimoles of `NaHCO_3` = 5x
Number of millimoles of `H_2CO_3` = 10 × 2 = 20
For the acidic buffer, `pH = pK_a + "log"([NaHCO_3])/([H_2CO_3])`
`7.4=-log(7.8xx10^(-7))+ log (5x)/20`
On solving, x = 78.37 ml
`therefore` Volume of `NaHCO_3` solution required = 78.37 ml.