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 should be mixed with a 10 mL sample of blood which is 2 Min `H_2 CO_3`, in order to maintain a pH of 7.4? `K_a` for `H_2 CO_3` in blood is `7.8 xx 10^(-7)`

Options

A. `50.20 mL`
B. `62.42 mL`
C. `78.37 mL`
D. `52.41 mL`

Answer

C

Explanation

Let us suppose the volume of `NaHCO_3` mixed be V mL
` therefore ` No. of moles of Na`HCO_3` in V mL of 5 M `NaHCO_3`
`=( 5 xx V )/(1000)= 0.005 V `
No of moles of` H_2 CO_3 ` in 10 mL of 2 M `H_2 CO_3 = ( 2 xx 10 )/( 1000 ) = 0.02`
Now from Henderson.s equation ,
`pH = pK_a + log (["salt "])/(["Acid"])`
`7.4 =- log ( 7.8 xx 10^(-7 ) ) + log ( 0.005 V )/( 0.02)`
`7.4 = 6.10 79 +log ( 0.005 V )/( 0.02)`
` 1.2921 = log ( 0.005 V )/( 0.02) or ( 0.005 V )/( 0.02 ) = 19.5929`
` therefore V= 78.37 mL`