Question

At ordinary body temperature (37c) the solubility of N2 in water that is in contact with air at ordinary atmospheric presure (1.00 atm) is 0.015g/L. Air is approximately 78 mol % N2.

A) calculate the molarity of dissolved N2 in blood, which can be treated as an aqueous solution, under these conditions.

B) Determine the Henry's law constant for N2 in blood (water) at 37 degrees C

C) Determine the solubility of nitrogen in blood at a depth of 100 ft. where the pressure is 4.0 atm.

D) if a scuba diver suddenly surfaces from the depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Answer 1

A)
P = 1 atm
p due to N2 = mole fraction of N2 * P
      = 0.78*1
       = 0.78 atm
At 1 atm, solubility = 0.015 g/L
p = KH*C
1 = KH * 0.015 g/L
KH = 66.7 atmL/g

So at 0.78 atm, it will be
C = p/KH
    = 0.78/66.7
    =0.0117 g/L
Molar mass of N2 = 14 g/mol
C = 0.0117 g/L
   =0.0117 g/L / 14 g/mol
    = 0.0117/14 mol/L
    = 8.36*10^-4 M
Answer: 8.36*10^-4 M

b)
As found above,
KH = 66.7 atmL/g
       = 66.7 atmL/g * 28 g/mol
      = 1867 atm-L/mol
Answer: 1867 atm-L/mol

c)
p due to N2 = 0.78*4 = 3.12 atm
C= p/KH
    =3.12 atm /1867 atm-L/mol
    = 1.67*10^-3 M

d)
AT depth,
C = 1.67*10^-3 M
At surface ,
C=8.36*10^-4 M
Moles of N2 released = 1.67*10^-3 - 8.36*10^-4 = 8.34*10^-4 mol
It is equal to 8.34*10^-4 *22.4 L = 0.0187 L = 18.7 mL
Answer: 18.7 mL