Chemistry 122 [Tyvoll]
Fall 2007
Practice Problems Using Henry’s Law (not to be turned in)
1. If 27 g of acetylene, C2H2, dissolves in 1.00 L of acetone at 1.0 atm pressure, (a) what
is the Henry’s law constant and (b) what is its solubility in acetone if the partial
pressure of C2H2 is 12.5 atm?
Solution. Begin with Henry’s Law: Sgas = KHPgas
(a) Substituting, 27.0 g C2H2/1.00 L = 27.0 g/L = KH(1.0 atm)
so, KH = (27.0 g/L)/( 1.0 atm) = 27.0 g/Latm
(b) Sgas = KHPgas = (27.0 g/Latm)(12.5 atm) = 337.5 = 338. g/L
2. The solubility of CO2 in water is 0.161 g/100 mL at 20 0C and a partial pressure of
CO2 of 760 mmHg. What partial pressure of CO2 is necessary in a soft drink canning
process in order to allow the solubility of CO2 to equal 0.886 g/100 mL?
Solution. Again, begin with Henry’s Law: Sgas = KHPgas
(a) Sgas = KHPgas
Since we have two sets of “conditions,” we can write S1/S2 = (KHP1/ KHP2)
or, canceling KH, S1/S2 = (P1/P2) where, here, S2 > S1, so P2 > P1
@ (0.161 g/100 mL)/( 0.886 g/100 mL) = (760 mmHg)/(P2)
or, P2 = (760 mmHg){(0.886 g/100 mL)/(0.161 g/100 mL)} = 4182 mmHg
P2 = 4.18 x 103 mmHg