Question

In water at 68°F at sea level (1 atm), water contains about 42 mg/L of dissolved oxygen. Determine the value of Henry's law constant for oxygen dissolving in water. Recall that air is 21% oxygen.

A) | 6.3 ´ 10–3 mol/(L atm) |

B) | 0.010 mol/(L atm) |

C) | 0.25 mol/(L atm) |

D) | 1.9 ´ 10–2 mol/(L atm) |

E) | 1.3 ´ 10–3 mol/(L atm) |

Answer #1

Recall that Henry's Law models solubility of GASES in liquids. Typically, as Pressure in the gas phase increases, the solubility of the gas increases. As Temperature increases, the solubility of gas will decrease ( contrary to salts)

Note that, for Henry's law to be applicable, the solute must be very low, i.e. << 0.1 (mol fraction)

S = Ki*Pi

Solubility of Gas in mol per liter = S

Hi = Henrys constant

Pi = Partial pressure of i

then, use:

S = mol of O2/L = 42 mg/32 = 1.3125 mmol/L = 1.3125*10^-3 mol/L

P = P-O2 = 0.21*1 atm = 0.21 atm of O2

then

S = H*P

H = S/P = (1.3125*10^-3)/0.21

H = 0.00625

H =6.25*10^-3 mol/L - atm

choose A

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