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Henry law constant for oxygen dissolved ...

Henry law constant for oxygen dissolved in water is` 4.34 xx 10^(4) atm` at `25^(@) C`. If the partial pressure of oxygen in air is 0.4 atm.Calculate the concentration ( in moles per litre) of the dissolved oxygen in equilbrium with air at `25^@C`.

Text Solution

Verified by Experts

Given :
Henry's law constant,`K_(H) =4.34 xx 10^(4) atm`
`P_(O_2)=0.4 atm`
According to Henry's law'
`p=K_(H)chi`
`:. pO_(2) = K_Hchio_2`
or `chio_(2)/K_(H)=0.4/4.34 xx 10^(4) =9.2 xx 10^(-6)`
Moles of water `(n_(H_(2)O)) = 1000/18 =55.5 mol`
Mole fraction of oxygen `(chiO_2)=(n_(O_2))/((n_(O_2))+n_(H_(2)O))`
Since `n_(O_2)` is very small in comparison to `n_(H_(2)O)`,
`:. chi_(O2)=n_(O_(2))/n_(H_(2)O) `
or `chi_(O_2) xx n_(H_2 O) = n_(O_2)`
`9.2 xx 10^(-6) xx 55.5 = n_(O_2)`
or `n_(O_2)=5.11 xx 10^(-4) mol`
Since `5.11 xx 10^(-4) mol` are present in `1000 mL` of solution, therefore , molarity = `5.11 xx 10^(-4)M`.
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