Question

(8 pts) Carbon dioxide in the atmosphere dissolves in water to establish an equilibrium that can increase the acidity of aqueous solutions in the environment.

This equilibrium is

CO2 (g) ⇌ CO2 (aq), Kh = 3.1 * 10^2 at 25°C.

Kh is called the Henry’s law constant, which relates the solubility of the gas CO2 in the aqueous solution, [CO2], to the partial pressure of CO2 over the solution, p(CO2):

[CO2] (M) = Kh * p(CO2) (atm)

During the preceding decades, the atmospheric concentration of CO2(g) has steadily increased, according to measurements at NOAA’s Mauna Loa Observatory in Hawaii.

(a) (2 pts) In Dec 2017, the mole fraction of CO2(g) was 405 ppm. What is the solubility (in M) of CO2(aq) in water at 25°C? You may assume the solution is in contact with air at 1.00 atm total pressure.

(b) (6 pts) Assume that all of this CO2(aq) is in the form of carbonic acid, H2CO3(aq). Determine the pH of this aqueous solution by setting up an I-C-E table. You may assume that H2CO3(aq) is just a monoprotic acid. Useful information is available in Appendix 5 of the textbook. (HINT: you should set up and solve a quadratic equation. – see Appendix 1 of the textbook)

NOTE: the pH of the oceans has dropped approximately 0.1 pH units from 8.2 to 8.1 since the start of the industrial revolution, over 200 years ago. The pH values of freshwater lakes and ponds vary from 6-8, depending on the surrounding soil and bedrock, which contain carbonates and other basic salts.

Answer 1

In this problem, we are going to use two important formulas

1) Partial pressure of CO₂ (Pan_CO2) = mol fraction of CO₂ * total pressure of the system

2) molar concentration of CO₂ = partial pressure of CO₂ (P_CO2) * K_H (given)

The given concentration of CO₂ is 405 ppm. So, molar fraction of CO₂ is = (405/10⁶) = 405*10^-6 [ ppm= parts per million, 1 million = 10⁶]

K_H of CO₂ at 25⁰C = 3.1*10^-2 mol/(L.atm) [Given data is 3.1*10² which is not correct. Reference: Wikipedia]

Now, we can consider the whole ocean as a system and the total pressure is 1 atm in this case. So, partial pressure of CO₂ = (405*10^-6*1) atm = 405*10^-6 atm.

The molar concentration of CO₂ = partial pressure of CO₂ * K_H = (405*10^-6) * (3.1*10^-2) {atm*mol/(L.atm)}= 1.256*10^-5 mol/L

Now, this 1.256*10^-5 mol/L of CO₂ gets dissolved in the sea water and produces H₂CO₃ i.e. carbonic acid

The chemical reaction is H₂O + CO₂ = H₂CO₃ [ 1 mol of CO₂ will produce 1 mol of H₂CO₃]

	H2O	CO2	H2CO3
Initial concentration (L/atm)	1.256*10-5	1.256*10-5	0
Change in concentration (L/atm)	(1.256*10-5-1.256*10-5) = 0	(1.256*10-5-1.256*10-5) = 0	(0+1.256*10-5) = 1.256*10-5
concentration in equilibrium (L/atm)	0	0	1.256*10-5

So, the concentration of H₂CO₃ in the sea water is 1.256*10^-5 L/atm. We get 2 mol of H⁺ ion from 1 mol of H₂CO₃. So, obtained H⁺ concentration from H₂CO₃ is (1.256*10^-5*2) = 2.512*10^-5 mol/L

As, the concentration of H₂CO₃ concentration is very low, we have to consider the H⁺ ions generated by the self-ionization of water. We consider that x mol of water molecules ionizes in 1 litre of water at 25⁰C. So, x mol of H⁺ and OH^- ions are generated in the water.

By forming the quadratic equation, we get

(x+2.512*10^-5) * x = 10^-14 [as the multiplication of H⁺and OH^-is always 10^-14 ]

by solving this equation using fx-991ES plus calculator, we get x = 3.98*10^-10

So, the total concentration of H⁺ ion is = (3.98*10^-10+ 2.512*10^-5) = 2.5120398*10^-5 mol/L

The definition of pH is the negative logarithmic value of molar concentration of H⁺

So, pH of the sea water = -log (2.5120398*10^-5) = 4.6