Question

The aqueous solubility of CO2 at 20 ℃ and 1.00 atm is equivalent to 87.8 mLCO2(g), measured at STP, per 100 mL of water.

Part A

What is the molarity of CO2 in water that is at 20 ℃ and saturated with air at 1.00 atm? The volume percent of CO2 in air is 0.0360%. Assume that the volume of the water does not change when it becomes saturated with air. Express your answer with the appropriate units.

 

Answer 1

Ans. It seems that there has been some type error in the question if you want a desired answer of “1.41 x 10^-5 M”.

Its assumed that the first line of question says “The aqueous solubility of CO₂ at 20.0⁰C and 1.00 atm is equivalent to 87.8 mL AIR, measured at STP, per 100.0 mL of water”.

If CO₂ solubility were actually 87.8 mL / 100 mL water (which is too high), there was no need of using the term “equivalent”.

# So, we have,

            Solubility of air at STP (1.00 atm, 0.0⁰C) = 87.8 mL

            Temperature at STP = 0.0⁰C = 273.15 K

Specified temperature, T = 20.0⁰C = 293.15 K

Calculate the volume of 87.8 mL air at 20.0⁰C using Charles’s Law as follow-

            V₁ / T₁(STP) = V₂ / T₂(20.0⁰C)

            Or, 87.8 mL / 273.15 K = V2 / 293.15 K

            Or, V2 = (87.8 mL / 273.15 K) x 293.15 K = 94.229 mL

Therefore, volume of air at 20.0⁰C = 94.229 mL

# Calculate volume of CO2 in air-

            Volume of CO2 in air = 0.0360 % of 94.229 mL

                                                = (0.0360 / 100) x 94.229 mL

                                                = 0.0339224 mL

                                                = 0.0000339224 L

# Calculate number of moles of CO₂

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation-     

            1.0 atm x 0.0000339224 L = n x (0.0821 atm L mol^-1K^-1) x 293.15 K

            Or, n = 1.4095 x 10^-6 mol

Hence, moles of CO₂ dissolved in 100.0 mL water = 1.4095 x 10^-6 mol

# Calculate Molarity-

            Molarity of CO₂ = Moles of CO₂ / Volume of solution in liters

                                    = (1.4095 x 10^-6 mol) / 0.100 L                               ; [100 mL = 0.100 L]

                                    = 1.4095 x 10^-5 mol/ L                                             ; [1 mol/ L = 1 M]

                                    = 1.4095 x 10^-5 M   =   1.41 x 10^-5 M