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
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