Question

a. Find Henry's constant of O2, N2, CO2, and He.
b. Order the solubilities of these gases in water in ascending order. Explain your answer.
c. Say how you could answer part b of this exercise without doing any computation.
d. Determine the pressure, in units of atmospheres (atm) necessary for the solubility of each gas in water to be 0.022 M?

Answer 1

a) KH for O2, N2, and He are 34.86. 76.48, and 144.97 katm respectively at 293K. KH for CO2 = 1.67kbar at 298K.

b) Solubility order = He<N2<O2<CO2.

c) p = KH.x

x= p/KH
Here KH is Henry’s law constant. x denotes mole fraction of the gas in solution

Thus KH is inversely proportional to mole fraction of gas in solution (representing its solubility). It is obvious from equation that higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. Hence, solubility order will be He<N2<O2<CO2.

d) Now, we know the KH of each gas. To find out the pressure of each gas necessary, we should have the mole fraction of each gas but the concentration is given in terms of molarity.So. first of all, we ll find out the mole fraction of each gas.

As we know,

x = n2/n1+n2

Let us find out the mol fraction of each gas, because solution is prepared in aqueous medium and conc. is 0.022 moles per liter.

nH2O = 1000/18 = 55.55 (1000ml = 1000g for water bcoz density = 1g/ml)

xHe = 0.022/0.022+55.55 = 3.96

xN2 = 0.022/0.022+55.55 = 3.96

likewise, for O2 and CO2 will be same.

Hence by putting the value of KH and mole fraction (x), we can find out the pressure,

pHe = 144.97x3.96 = 574.08 katm

pN2 = 76.48x3.96 = 302.86 katm

pO2 = 34.86x3.96= 138.04 katm

pCO2 = 1.67x3.96= 6.6 katm