In natural water systems, carbon dioxide gas dissolves in the water to form carb

Question

In natural water systems, carbon dioxide gas dissolves in the water to form carbonic acid, H_2 CO_3. a) Write the chemical reaction for this process. b) Estimate the equilibrium constant, K, for this reaction at 25 degree C Carbonic acid then protonates the water in which it is dissolved. Estimate the equilibrium constant, K for the overall process by which gaseous carbon dioxide and water fork hydrodium ion, H_3O^+, and bicarbonate, HCO_3^-(One can ignore the second protonation step) What is the pressure of CO_2 in equilibrium with carbonated water at 25 degree C and pH = 3.60?

Explanation / Answer

Answer to Qa)

CO2 (g) + H2O (l) ------> H2CO3 (aq)

Answer to Qb)

We know,

G = G0 +RT lnQ

The Gibbs free energy change is given as:

G0 = [G0 (H2CO3(aq))] – [G0 (CO2 (g)) + G0 (H2O (l))]

Using G0f values from the thermodynamic table

G0 = (-623.4) - [(-394.4) + (-237.1)]

= -623.4 + 631.5

= 8.1 kJ

At equilibrium G = 0 and Q = Ksp

G = G0 +RT ln Q

0 = 8.1 + 0.00831 X 298 ln Ksp

ln Ksp = -8.1 / (0.00831 X 298)

= -3.27

Ksp = e-3.27

= 3.8 X 10-2

Note Ksp does not have units

Answer to Qc)

This reaction for protonation of water can be given as:

CO2 (g) + H2O (l) ------> H3O+(aq) + HCO3-(aq)

The Gibbs free energy change is given as:

G0 = [G0 (HCO3-(aq))+ G0 (H3O+(aq)) ] – [G0 (CO2 (g)) + G0 (H2O (l))]

Using G0f values from the thermodynamic table

G0 = [(-587.1) + (-237.1)] - [(-394.4) + (-237.1)]

Note, G0 (H3O+(aq)) value will be same as G0 (H2O (l)) since the G0 for addition of a proton is zero

= -824.2 + 631.5

= -192.7 kJ

At equilibrium G = 0 and Q = Ksp

G = G0 +RT ln Q

0 = -192.7 + 0.00831 X 298 ln Ksp

ln Ksp = 192.7 / (0.00831 X 298)

= 77.82

Ksp = e77.81

= 6.23 X 1033

Answer to Qd)

pH = -log [H+]

[H+] = antilog(-pH)

=antilog (- 3.60)

[H+] = 3 X 10-4 M

Carbonic acid dissociates as:

H2CO3 (aq)------> H+(aq) + HCO3-(aq)

[HCO3-] = [H+] = 3 X 10-4 M

Ka for carbonic acid = 4.5 X 10-7

Ka = [HCO3-] X [H+] / [H2CO3]

[H2CO3] = [HCO3-] X [H+] / Ka

= (3 X 10-4 )2 / 4.5 X 10-7

= 9 X 10-8 / 4.5 X 10-7

= 0.2 M

CO2 (g) + H2O (l) ------> H2CO3 (aq)

[CO2] = [H2CO3] = 0.2 M

Henry's law states, “At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.” Mathematically this law is given as:

p = kH•c

where, p = partial pressure of the solute in a solution, c = concentration of the solute in a solution and kH = Henry’s Law constant.

p = kH•c

= 29 atm M-1•0.2 M = 5.8 atm

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