Question

a)

COAST Tutorial Problem The solubility of air in water is approximately 2.1 x103 M at 20°C and 1.0 atm. Calculate the Henry's law constant for air Number mol/ L atm Is the KH value of air approximately equal to the sum of the KH values of N2 and O2 because these two gases make up 99% of the gases in air? O Yes O No

b)

c)

Answer 1

Solution :-

1). Solubility of air = 2.1*10^-3 M at 20 c and 1 atm

KH = C/P

   = 2.1*10^-3 M / 1 atm

    = 2.1*10^-3 mol / L . atm

So the henrys law constant for Air = 2.1*10^-3 mol / L .atm

The Henrys law constat for the O2 is 1.28*10^-3 mol per L atm

So it is Yes .

a). Solubility of O2 = 6.5 mg / L

6.5 mg * 1 g / 1000 mg = 0.0065 g

0.0065 g * 1 mol / 32 g = 0.000203 mol

So the solubility of the O2 = 2.03*10^-4 mol / L

Now lets calculate the constant

KH= c/p

     = 2.03*10^-4 mol / L / 1 atm

     = 2.03*10^-4 mol / L . atm

b) pressure of CO2 = 46 psi * 1 atm / 14.7 psi = 3.13 atm

KH for CO2 = 0.037 mol per L atm

Now lets calculate the solubility of the CO2

KH= c/p

C= KH* P

= 0.037 mol per L atm * 3.13 atm

=0.1158 mol / L