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

`[H_(2)CO_(3)]` in blood = 2 M
Volume of blood = 10 mL
`[NaHCO_(3)]=5M`
Let volume of `NaHCO_(3)` used = V mL
`:. [H_(2)CO_(3)]` in mixture `=(2xx10)/((V+10))`
`[NaHCO_(3)]` in mixture `= ((5+V))/((V+10))`
`:. pH=pK_(a)+log ((["Salt"])/(["Acid"]))`
`7.4= -log 7.8xx10^(-7) + log (((5xxV)//(V+10))/((2xx10)//(V+10)))`
`:. V = 78.36 mL`