Question

Calculate the solubility of H2 in water at 25 degree celsius if its partial pressure above the solution is 1 bar. given that henry's constant for H2 in water at 25 degree celsius is 71.18 kbar.

need the answer very urgentlyy!!

Answer 1

Henry's law states that :

$p = k_H\ c$

where p = partial pressure of the gas above the solution.
    $k_H$ is the Henry's solubility constant for the solvent & solute at given T°C
    c = molar concentration  of the solute in the solution in  moles/Litre  at given T°C

p = 1 bar    and    $k_H$=71.18 kbar-Litre/mol
   c = 1  / 71.18 * 10³ = 14.048 mol/Litre

 concentration =  14.048 mol/Litre  or    28.096 gms/Litre

Answer 2

ANSWER :

$7.79 \times {10}^{ - 4}$

EXPLANATION :

according to Henry's law:-

pH2=kHxH2

where,pH2=partial pressure of h2

kH =henry’s constant

xH2=mole fraction of h2.

pH2= KH2 X H2

XH2=PH/ KH

=1 bar / 71.18×103 bar

=1.404×10-5

now,

nH2O in 1 L=1000g/ 18=55.55

therefore,

nH2/ 55.55=1.404×10-5

nH2= 7.79×10-4 moles

hence the solubility of h2 in water is 7.8×10-4 mol/l