henry law constant for oxygen dissolved in water is 4.34 into 10 to the power 4 atmosphere at 25 degree Celsius the partial pressure of oxygen in air is 0.2 ATM and ordinary atmospheric condition calculate the concentration of dissolved oxygen in water in
equilibrium with air at 25 degree Celsius
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According to Henry's law, P=K
H
×x
∴x
O
2
=
K
H
P
O
2
=
4.34×10
4
0.2
=4.6×10
−6
Moles of water=
18
1000
=55.5mol
∴x
O
2
=
n
H
2
O
+n
O
2
n
O
2
≈
n
H
2
O
n
O
2
⇒n
O
2
=4.6×10
−6
×55.5=2.55×10
−4
mole
∴ Molarity = 2.55×10
−4
M
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