At what partial pressure, oxygen will have solubility of 0.05 g L-1 in water at 293 K? Henry’s constant (KH) for O2 in water at 293 K is 34.36 k bar. Assume the density of the solution to be same as that of the solvent.

At what partial pressure, oxygen will have solubility of 0.05 g L-1 in water at 293 K? Henry’s constant (KH) for O2 in water at 293 K is 34.36 k bar. Assume the density of the solution to be same as that of the solvent.

Answer:

Given Values

A = H2O    ,    B = O2        

KH = 34.36 k bar = 34.36 × 103 bar

Solubility = 0.05 gm/L

∴ WB = 0.05 gm , Vsol = 1 L or 1000 ml

MB = 2(16) = 32

Dsol = Dwater (A) = 1 gm/ml

∴ Wsol = 1000 gm

WA = Wsol − WB  = 1000 − 0.005 ≈ 1000 gm

MA = 2+16 = 18

Calculations

Calculate mole of Oxygen (nB)

\( n_{B} = \frac{W_{B}}{M_{B}} = \frac{0.05}{32} = 0.0016 \) moles

Calculate mole of Solvent or Water (nA)

\( n_{A} = \frac{W_{A}}{M_{A}} = \frac{1000}{18} = 55.5 \) moles

Calculate Partial pressure of Oxygen

\( P = K_{H}\frac{n_{B}}{n_{A}+n_{B}} \)

\( P = 34.36 \times 10^{3}~ \frac{0.0016}{55.5+0.0016} \)

P = 0962 bar

You should check out our other content to boost your self-study.

  • You can read and download our notes
  • You can practice multiple choice questions.
  • You can practice short and long answer questions
  • You can watch our pre-recorded video lectures. 
  • You can participate in multiple choice questions quiz.