Four liters of water are equilibrated with a gas mixture containing carbon dioxi
ID: 792201 • Letter: F
Question
Four liters of water are equilibrated with a gas mixture containing carbon dioxide at a partial pressure of .3 atm. Henry's law constant for H2CO3* solubility is 2.0 g/L-atm.
a. How many grams of Co2 are dissolved in the water? What is the pH of this water? (Neglect the secondary dissociation of carbonic acid.)
b. Assume that H2CO3* dissolves first without dissociating, achieving equilibrium with the atmosphere, and then the solution is closed to the atmosphere. H2CO3* then dissociates (neglect the secondary dissociation). Find the PH of this solution and compare it to the result obtained in (a).
Explanation / Answer
p=kh*c
P=0.3atm
Kh = 62/2.0 atm/M = 31.0atm/M [given constant has units of g/l*atm. Actual constant has unitsL*atm/mol]
[Mol. wt. of carbonic acid = 62g/mol]
Concentration "C" of H2CO3 in water-gas equilibriated mixture = 0.3/31 M = 0.0097 M
Moles of CO2 dissolved = 0.0097*4 =0.039 M
Weight of CO2 dissolved = 0.039*44=1.703g
now, H2CO3 dissociates.
H2CO3<=> H+ + HCO3-
0.0097-x x x
ka = x^2/(0.0097-x)
2.5*10^(-4) = x^2/(0.0097-x)
[H+] = 1.437*10^(-3) M
pH = -log[H+] = 2.842
b) 0.04% is the concentration of CO2 in atmosphere
Partial pressure of CO2 = 0.0004atm
.Concentration of H2CO3 = 0.0004/31 = 1.3*10^(-5) M
H2CO3 <=> H+ + CO3-
1.3*10^(-5)-x x x
2.5*10^(-4) = x^2/(1.3*10^(-5)-x)
x= 1.2386*10^(-5)
pH = -log[H+] = 4.907
Since, the partial pressure of CO2 is lower in atmosphere, the pH is comparatively higher than the gas mixture used for aeration.