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PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE     223

                                     Table 5.4  Henry’s law constants
                                                          ◦
                                     k H for gases in water at 25 C
                                     Gas        k H /mol dm −3  bar −1
                                     CO 2          3.38 × 10 −2
                                     O 2           1.28 × 10 −3
                                     CH 4          1.34 × 10 −3
                                     N 2           6.48 × 10 −4



             partial pressure) of the gas above it. Stated in another form, Henry’s
             law says:                                                    One of the simplest
                                                                          ways of removing
                                                                  (5.20)  gaseous oxygen from
                                    [i (soln) ] = k H p i
                                                                          water is to bubble
             where p i is the partial pressure of the gas i, and [i (soln) ]isthe  nitrogen gas through
             concentration of the material i in solution. The constant of propor-  it (a process called
             tionality k H is the respective value of Henry’s constant for the gas,  ‘sparging’).
             which relates to the solubility of the gas in the medium of choice.
             Table 5.4 lists a few Henry’s law constants, which relate to the
             solubility of gases in water.
                                                                          Strictly, Henry’s law
             Worked Example 5.7 What is the concentration of molecular oxy-  only holds for dilute
                            ◦
             gen in water at 25 C? The atmosphere above the water has a pressure  systems, typically in
                 5
             of 10 Pa and contains 21 per cent of oxygen.                 the mole-fraction range
                                                                          0–2 per cent. The law
             Strategy. (1) We calculate the partial pressure of oxygen p (O 2 ) .(2) We  tends to break down
             calculate the concentration [O 2(aq) ] using Henry’s law, Equation  as the mole fraction
             (5.20), [O 2(aq) ] = p (O 2 ) × k H(O 2 ) .                  xincreases.

               (1) From the partial of oxygen p (O 2 ) = x (O 2 ) × the total pressure p (total) ,where x is
                    the mole fraction:
                                                       5
                                        p (O 2 ) = 0.21 × 10 Pa
                                                      4
                                        p (O 2 ) = 2.1 × 10 Pa or 0.21 bar
               (2) To obtain the concentration of oxygen, we insert values into Henry’s law,
                    Equation (5.20):
                                                     O
                                     [O 2(aq) ] = 0.21 × p × 1.28 × 10 −3  mol dm −3  bar −1
                                     [O 2(aq) ] = 2.69 × 10 −4  mol dm −3
                                                                          This relatively high
                                                                          concentration of oxy-
               We need to be aware that k H is an equilibrium constant, so
                                                                    ◦
             its value depends strongly on temperature. For example, at 35 C,  gen helps explain why
                                                                          fish can survive in
             water only accommodates 7.03 mg of oxygen per litre, which ex-
                                                                          water.
             plains why fish in warm water sometimes die from oxygen
             starvation.
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