Question

The solubility of carbon dioxide in water is of considerable environmental and commercial importance. Although the behavior of dissolved CO2 is somewhat complicated, for most purposes it can be treated as dissolving to form carbonic acid, H2CO3, according to the following equation: CO2 (g) + H2O (l) H2CO3 (aq) Keq = 0.036

(a) Carbonated sodas are typically packaged under a carbon dioxide pressure of about 5 bar. Calculate the number of moles of carbon dioxide dissolved (in the form of H2CO3) in a 2.00-L bottle of soda.

(b) The headspace above the liquid in a typical 2-L bottle of soda has a volume of about 50 mL. Estimate the number of moles of CO2 that are in the headspace of the soda bottle before it is opened by the consumer.

(c) A commercial device intended to prevent soft drinks in opened bottles from going flat (losing their dissolved CO2) is called the “Fizz-Keeper.” The Fizz-Keeper operates by screwing on to the top of the opened bottle and pumping air into the bottle to restore its original pressure. Would this decrease the loss of carbon dioxide from the solution, compared to just screwing the original cap back on firmly? Explain your reasoning.

(d) An internet source (http://www.lenntech.com/carbon-dioxide.htm) states that the dissolution of carbon dioxide from the air “explains why water, which normally has a neutral pH of 7 has an acidic pH of approximately 5.5 when it has been exposed to air.” Assess the validity of this claim. Air consists of about 0.04% by volume of carbon dioxide (400 ppm); the pKa of H2CO3 is 6.37. (2) If water is pure enough and in a sufficiently smooth contai

Answer 1

(a) Carbonated sodas are typically packaged under a carbon dioxide pressure of about 5 bar.
Calculate the number of moles of carbon dioxide dissolved (in the form of H2CO3) in a 2.00-L bottle of soda.

pCO2 = 5bar
1.013 bar = 1 atm
pCO2 = 5 bar = 5/1.013 = 4.936 atm

CO2(g) + H2O(l) = H2CO3(aq); Keq = 0.036
Keq = [H2CO3]/pCO2
[H2CO3] = Keq*pCO2 = 0.036*4.936 = 0.1777 M
Volume = 2 L
no of moles of H2CO3, n = 0.1777*2 = 0.355 moles

(b) The headspace above the liquid in a typical 2-L bottle of soda has a volume of about 50 mL.
Estimate the number of moles of CO2 that are in the headspace of the soda bottle before it is opened by the consumer.

Let the bottle is at 25C = 298 K
Volume of head space, V = 50 ml = 0.05 L
1.013 bar = 1 atm
P = 5 bar = 5/1.013 = 4.936 atm
Gas constant, R = 0.0821 L atm/mol K
no of moles in head space, n = P*V/(R*T) = 4.936*0.05/(0.0821*298) = 0.01 mol

(c) A commercial device intended to prevent soft drinks in opened bottles from going flat (losing their dissolved CO2) is called the “Fizz-

Keeper.” The Fizz-Keeper operates by screwing on to the top of the opened bottle and pumping air into the bottle to restore its original pressure.

Would this decrease the loss of carbon dioxide from the solution, compared to just screwing the original cap back on firmly? Explain your

reasoning.

This will not change the loss of CO2.

CO2(g) + H2O(l) = H2CO3(aq); Keq = [H2CO3]/pCO2
No of moles of CO2(g) will remain same with or without adding inert air.
This will make pCO2 does not change at higher pressure, because total pressure will increase and mole fraction will decrease.
So with pCO2 remaining same, [HCO3] = Keq*pCO2 will remain same and no change will happen to the equilibrium concentration.

(d)
(i)An internet source (http://www.lenntech.com/carbon-dioxide.htm) states that the dissolution of carbon dioxide from the air “explains why water,

which normally has a neutral pH of 7 has an acidic pH of approximately 5.5 when it has been exposed to air.” Assess the validity of this claim.

Air consists of about 0.04% by volume of carbon dioxide (400 ppm); the pKa of H2CO3 is 6.37.

Mole fraction of CO2 in air, yCO2 = 0.04/100
Total pressure, Pt = 1atm
pCO2 = yCO2*Pt = 0.0004 atm

CO2(g) + H2O(l) = H2CO3(aq); Keq = 0.036
Keq = [H2CO3]/pCO2
[H2CO3] = Keq*pCO2 = 0.036*0.0004 = 0.0000144 M

pKa = 6.37
Ka = 10^(-6.37)

H2CO3(aq) = HCO3(-)(aq) + H+(aq); Ka = 10^(-6.37)
Ka = [HCO3(-)]*[H+]/[H2CO3]

Initial concentration:
[HCO3(-)] = 0, [H+] = 0, [H2CO3] = 0.0000144 M

Change:
[HCO3(-)] = x, [H+] = x, [H2CO3] = -x M

Equilibrium concentrations:
[HCO3(-)] = x, [H+] = x, [H2CO3] = 0.0000144 - x M

Ka = [HCO3(-)]*[H+]/[H2CO3] = x*x/( 0.0000144 - x) = 10^(-6.37)
[H+] = x = 2.5E-6
pH = -Log(H+) = 5.6

This is matching the observed pH of 5.5