CHEMISTRY 253

Spring, 2015 - Dixon

Homework Set 1.5 Solutions – Collected Problems

 

Problems:

4-3  An air-filtering device is tested and found to remove all particles larger than 1 mm in diameter, but almost none of the smaller ones.  Calculate the percentage of surface area removed by the device for a sample of particulates, 95% of the mass of which is particles of diameter 10 mm and 5% of which is particles of 0.1 mm.  Assume all particles are spherical and of equal density.                                                                                                

Let’s call the total mass 1.  Then the mass of the 10 mm particles is 0.95 and of the 0.1 mm particles is 0.05.  The number of particles is given by dividing the mass by the volume of each particle (4pr3/3)

10 mm particles: (0.95)/[(4)(3.1416)(5 mm)3/3] = 0.001814

0.1 mm particles: (0.05)/[(4)(3.1416)(0.05 mm)3/3] = 95.5

(not worrying about units yet – just that they are the same for each particle type)

Now to convert to surface area = N*4pr2

10 mm particles: surface area = (0.001814)(4)(3.1416)(5 mm)2 = 0.570

0.1 mm particles: surface area = (95)(4)(3.1416)(0.05 mm)2 = 3.00

% surface area in 10 mm particles = 0.570*100/(0.570 + 3.00) = 16%

% surface area in 0.1 mm particles = 84% so 16% removed

 

Additional Problems:

4-1  A sample of acidic precipitation is found to have a pH of 4.2.  Upon analysis, it is found to have a total sulfur concentration of 0.000010 M.  Calculate the concentration of nitric acid in the sample, and from the ratio of nitric to total acid, decide whether the air sample probably originated in eastern or in western North America.

[Note: I’m making the assumption that total sulfur is from sulfuric acid and not from sulfite and without any ammonium – which is a bad assumption.  This gives the minimum sulfate concentration].

If [H2SO4] = 1.0 x 10-5 M, then [H+](from H2SO4) = 2.0 x 10-5 M

From the pH, the total [H+] = 10-4.2 = 6.3 x 10-5 M

[H+](from HNO3) = total [H+] - [H+](from H2SO4) = 6.3 x 10-5 M - 2.0 x 10-5 M = 4.3 x 10-5 M

With the assumptions made, it should be from western North America since [HNO3] > [H2SO4].  In reality, without information about ammonium, it is impossible to say.

 

4-3  The pH in a lake of size 3.0 km x 8.0 km and average depth of 100 m is found to be 4.5.  Calculate the mass of calcium carbonate that must be added to the lake water in order to raise its pH to 6.0.

n(H+) in lake = [H+]V = (10-4.5)(3000 m)(8000 m)(100 m)(1 L/0.001 m3) = 7.589 x 107 mol

n(H+) in “neutralized” lake = (10-6.0)(3000 m)(8000 m)(100 m)(1 L/0.001 m3) = 2.40 x 106 mol

n(H+) needed = 7.589 x 107 mol - 2.40 x 106 mol = 7.349 mol x 107 mol

From reaction, 2H+ + CO32- → CO2 + H2O, we see we need ˝ mol CO32- per mol of H+ or

n(CaCO3) needed = 0.5(7.349 mol x 107 mol) = 3.675 x 107 mol

mass needed = (100.1 g/mol)(3.675 x 107 mol) = 3.7 x 106 kg CaCO3