CHEMISTRY 253

Spring, 2015 - Dixon

Homework Set 1.4 Solutions

Problems to Turn In

Problems:

3-10 What mass of calcium carbonate is required to react with the sulfur dioxide that is produced by burning one tonne (1000 kg) of coal that contains 5.0% sulfur by mass?

One tonne of coal means n(S) =  (0.050)(1000 kg)(1000 g/kg)(1 mol/32.064 g) =  1559 mol S

Reaction: CaCO₃(s) + H₂SO₃(aq) → CaSO₄ (s) + CO₂ (g) + H₂O (l)

Mass CaCO₃ = (1559 mol S)(1 mol CaCO₃/1 mol S)(100.1 g/ mol CaCO₃)(1 kg/1000 g)

 = 156 kg CaCO₃

3-14  Calculate the concentration of SO₂ that must be reached in polluted air if the dissolved gas is to produce a pH of 4.0 in raindrops without any oxidation of the sulfur.

To keep this simple so that it can be solved, we can assume that little SO₂ dissolves in the rain and that the raindrop pH is totally due to the SO₂ dissolving (no acidity form other gases or aerosol particles).  Since pH << pK, we also can ignore formation of SO₃^2-.

Reactions:       SO₂ (g) + H₂O (l)  → H₂SO₃ (aq)

            and      H₂SO₃ (aq)  → H⁺ + HSO₃^-

K_H = [H₂SO₃ (aq)]/P_SO2 = 1.0 M/atm and K_a1 = 0.017 = [H⁺][HSO₃^-]/[H₂SO₃ (aq)]

and finally, [H⁺] = [HSO₃^-] = 10^-4.0

Combining these, [H₂SO₃ (aq)] = (10^-4.0)(10^-4.0)/(0.017) = 5.88 x 10^-7 M

and P_SO2 = (5.88 x 10^-7 M)/(1.0 M/atm) = 5.88 x 10^-7 atm

At sea level pressure (P_air = 1.00 atm), this corresponds to a mixing ratio of 0.59 ppmv

3-15  a) Confirm by calculation that the pH of CO₂-saturated water at 25ºC is 5.6, given that the CO₂ concentration in air is 390 ppm.  For carbon dioxide, the Henry’s law constant K_H = 3.4 x 10^-2 M atm^-1 at 25ºC.  The K_a1 for carbonic acid, H₂CO₃, is 4.5 x 10^-7 at that temperature.

K_H = [H₂CO₃ (aq)]/P_CO2 = 0.034 M/atm

so [H₂CO₃ (aq)] = (0.034 M/atm)(390 x 10^-6 atm) = 1.33 x 10^-5 M

K_a1 = 4.5 x 10^-7 = [H⁺][HCO₃^-]/[H₂CO₃ (aq)] = [H⁺]²/[H₂CO₃ (aq)]

or [H⁺] = [(4.5 x 10^-7)(1.33 x 10^-5)]^0.5 = 2.44 x 10^-6 M or pH = 5.61

b) Recalculate the pH for a carbon dioxide concentration of 560 ppm, i.e., double that of the preindustrial age.

[H₂CO₃ (aq)] = (0.034 M/atm)(560 x 10^-6 atm) = 1.90 x 10^-5 M

[H⁺] = [(4.5 x 10^-7)(1.33 x 10^-5)]^0.5 = 2.93 x 10^-6 M or pH = 5.53

3-18  Let k be a given measure of length; then suppose a cubic particle of dimensions 3k x 3k x 3k is split up into 27 cubes of size k x k x k.  Calculate the relative increase in surface area when this occurs by comparing the surface area (length times width of the six faces of the larger cube to the sum of all those of the smaller ones.  From your answer deduce whether the total surface area of a given mass of atmospheric particles is larger or smaller when it occurs as a large number of small particles rather than a small number of larger ones.

large cube: surface area = 6(3k)² = 54k (6 faces with dimensions 3k x 3k each)

small cube: surface area = 27(6)(k)² = 162k.

 So 162/54 = 3 times as much total surface area for the smaller cubes.

Additional Problems:

3-6  The percentage of sulfur in coal can be determined by burning a sample of the solid and passing the resulting sulfur dioxide gas into a solution of hydrogen peroxide, which oxidizes it to sulfuric acid, and then titrating the acid.  Calculate the mass percent of sulfur in a sample if the gas from an 8.05 g sample required 44.1 mL of 0.114 M NaOH in the titration of the diprotic acid.

n(H₂SO₄) = (1 mol H₂SO₄/2 mol OH^-)(44.1 mL)(0.114 mol/L)(L/1000 mL) = 0.002514 mol S

mass % = (g S)(100)/g coal = (0.002514 mol S)(32.064 g S/mol)(100)/8.05 g

mass % = 1.0%

3-9  Assuming its concentration in air is 2.0 ppb, calculate the molar solubility of SO₂ in raindrops whose pH is fixed (by presence of strong acids) to be 4.0, 5.0 and 6.0.  The data required for the calculations is present in Section 3.21 of the text.

It is not clear what is meant by molar solubility as solubility has to do with concentration per amount of substance being dissolved.  I’m assuming it is asking for molarity in equilibrium with the gas phase concentration (also assuming, very little SO is transferred to the gas phase).

At pH = 4.0

[SO₂]_dissolved = [H₂SO₃ (aq)] + [HSO₃^-] + [SO₃^2-]

[H₂SO₃ (aq)] = K_HP_O2 = (1.0 M/atm)(2 x 10^-9 atm) = 2 x 10^-9 M

K_a1 = [H⁺][HSO₃^-]/[H₂SO₃ (aq)] and K_a2 = [H⁺][SO₃^2-]/[HSO₃^-]

[HSO₃^-] = [H₂SO₃ (aq)]K_a1/[H⁺] = (2 x 10^-9)(0.017)/(10^-4.0) = 3.4 x 10^-7 M

[SO₃^2-] = [HSO₃^-]K_a2/[H⁺] = (3.4 x 10^-7)(1.2 x 10^-7)/(10^-4.0) = 4.1 x 10^-10 M

[SO₂]_dissolved = 2 x 10^-9 M + 3.4 x 10^-7 M + 4.1 x 10^-10 M = 3.4 x 10^-7 M

At pH = 5.0

[H₂SO₃ (aq)] = 2 x 10^-9 M (pH independent)

[HSO₃^-] = [H₂SO₃ (aq)]K_a1/[H⁺] = (2 x 10^-9 M)(0.017)/(10^-5.0) = 3.4 x 10^-6 M

[SO₃^2-] = [HSO₃^-]K_a2/[H⁺] = (3.4 x 10^-6)(1.2 x 10^-7)/(10^-5.0) = 4.1 x 10^-8 M

[SO₂]_dissolved = 2 x 10^-9 M + 3.4 x 10^-6 M + 4.1 x 10^-8 M =3.4 x 10^-6 M

At pH = 6.0

[HSO₃^-] = [H₂SO₃ (aq)]K_a1/[H⁺] = (2 x 10^-9 M)(0.017)/(10^-6.0) = 3.4 x 10^-5 M

[SO₃^2-] = [HSO₃^-]K_a2/[H⁺] = (3.4 x 10^-5)(1.2 x 10^-7)/(10^-6.0) = 4.1 x 10^-6 M

[SO₂]_dissolved = 2 x 10^-9 M + 3.4 x 10^-5 M + 4.1 x 10^-6 M =3.8 x 10^-5 M