CEE 370

Fall 2010

Homework #2

 

1.    1 pointHow many grams of anhydrous Sodium Sulfate (Na2SO4) must be added to a liter of water so that the ionic strength equals 0.01M?

 

 

for a solution of sodium sulfate at a molar concentration of “C”, we have “C” moles of sulfate (a divalent ion) and “2C” moles of sodium (a monovalent ion).  Therefore,

 

 

since the GFW of Na2SO4 is (2x23)+32+(4x16)=142

 

 

 

2.    1 pointThe pesticide, parathion, undergoes photolytic degradation (destruction by absorption of energy from the sun’s rays) in natural waters.  This is a first order process with a rate constant of 8x10-7s-1.

a.    What is the half-life for parathion based on this degradation process?

 

b.    How long will it take for 90% of the parathion to degrade?

 

c.    How long will it take for 99% of the parathion to degrade?

 

 

3.    1 pointHypochlorous acid (HOCl) reacts with phenol (C6H5OH) to produce a wide range of products including highly offensive chlorophenols.  The rate of this reaction is second order, overall (first order in each reactant).  The rate constant is 2x103 M-1s-1.  In this problem you have 1.2 ppb of phenol in water, and to this you add 0.1 mM hypochlorous acid.

a.    How long will it take to destroy 50% of the phenol?

 

First, calculate molar phenol concentration.

Although this is a second order kinetic problem, one of the reactants (hypochlorous acid) is present at a very large molar excess (e.g., 0.1 mM vs 0.0000128mM).  Therefore, this reactant will not become depleted as the reaction proceeds, and the overall reaction takes on the form of pseudo-first order kinetics.

 

So the following equation will hold:

Where:

and now we can simply used the re-arranged form of the first order kinetic equation as developed in question #2 above.

b.    How long will it take to destroy 99.99% of the phenol?

 

And using the same equation again:

 

 

4.    2 pointsCalculate the ThOD of the following wastewaters

a.    45 mg/L of methanol

The balanced stoichiometric equation is:

The ThOD is then:

 

b.    10 mg/L of glucose

The balanced stoichiometric equation is:

The ThOD is then:

 

 

c.    45 mg/L of methanol with 10 mg/L glucose

The ThOD of the two components are calculated in parts a and b.  The answer is the sum of the two, or:

 

d.    10-4 M acetic acid

First, the mass-based concentration is:

The balanced stoichiometric equation is:

The ThOD is then:

 

e.    5 mg/L of trichloroethane

The balanced stoichiometric equation is:

The ThOD is then:

 

5.    1 pointNitrogen occurs at a variety of oxidation states in water environments. Balance the following oxidation-reduction reaction which is important in wastewater treatment:

 

NH3 + O2 → NO3- + H2O 

 

Show all steps.

 

Oxidation:

Reduction:

 

Multiply the reduction reaction by 2x and adding the two we get: