CEE 670 |
Fall 2011 |

**26.** A glass vial contains 0.5%
potassium by weight. The natural
abundance of ^{40}K, which decays by β-emission is 0.00118%, and
its half-life is 1.28x10^{9} years.
If the vial weighs 15g, calculate the expected rate of β^{-}
production by the vial in disintegrations per minute or dpm. (problem 1 in Chapter 2 of Brezonik; pg.102).

__Amount present now (Mass or Atoms)__

15 g total weight

[^{40}K] =
15g*(0.5/100)*(0.00118/100) = 8.85 x 10^{-7} g

= 8.85 x 10^{-7} g ÷ 40 g/M = 2.21 x 10^{-8} M

=
6.028 x 10

**1 Point**

__Rate:__

T ½ = 1.28 x10^{9}
years = 6.73 x 10^{14} min

K= ln(2)/ t ½ = 0.693/ 6.73 x 10^{14} = 1.0296 x 10^{-15}
min^{-1}

Disintegrations = 1.33 x 10^{16}
atoms x 1.0296 x 10^{-15} min^{-1}

= 13.72
atoms/m or dpm

**27.** Using the kinetic plot and
data for the hydrolysis of Benzyl Chloride as presented in class (refer to posted slides), determine the following:

a.

The
expected concentration of benzyl chloride after 72 hours (@25°C), when the initial concentration was
33µg/L.

**1 Point**

C = __0.923 µg/L__

b. The expected concentration
of benzyl chloride after 72 hours (@25°C), when the initial
concentration was 7.9µg/L.

C = __0.221 µg/L__

c. The expected concentration
of benzyl chloride after 72 hours (@0.1°C), when the initial
concentration was 33µg/L.

C = __29.6 µg/L__

**28.** It has been known for some
time that concentrated solutions of aqueous chlorine lose strength with
time. Accompanying this is the
accumulation of chlorate ion. Early
chemical studies have determined this to be a second order reaction in hypochlorite. The stoichiometry is as follows:

Bolyard
and co-workers [1992, Env. Sci. Technol. 26(8)1663-1665] found that chlorate is
being inadvertently added to drinking water when aqueous chlorine is used as a
disinfectant. Data collected at 14 sites
showed that raw water chlorate levels were 0.02 mg/L or below. However, finished water levels were as high
as 0.66 mg/L, apparently due to contamination from the chlorine stocks. Gordon and Adam [Water Disinfection News,
5:1, Novatek] have subsequently studied this reaction. They determined that a 2.776 M chlorine stock
solution (15.89% FAC, d=1.239 g/mL) has a 61.7 day half-life at 25°C and pH 13.
Based on this information answer the following questions.

a. What is the 2^{nd}
order rate constant for this reaction in units of M^{-1}s^{-1}?

First recognize that all of the chlorine (2.776 M) is in the form of hypochlorite (OCl-) at pHs
greater than the pKa (7.6) and that chlorine is expressed as Cl2 (GFW = 71). Note also that the parenthetical note “(15.89%
FAC, d=1.239 g/mL)” is what I used to calculate the 2.776 M concentration. You don’t need to use this information in the
solution.

**1 Point**

For a half-life:

k =__2.25 x 10 ^{-8} M^{-1}s^{-1}__

b. Calculate the concentration
of chlorate in the stock after 40 days holding time at 25°C

First calcuate the final hypochlorite concentration

and

So at 40 days

C = [OCl-] = 1.68
M

And recognize that the chlorate formed is equal to one-third times
the hypochlorite lost

[ClO_{3}^{-}]
= (Co-C)/3

[ClO_{3}^{-}]
= __0.364 M__

c. What is the dosed
concentration (mg/L) of chlorate if sufficient stock (aged 40 days at 25°C) is added to a finished drinking water to
achieve a free chlorine dose of 2.5 mg/L?

For this you’ll want to calculate the molar ratio of chlorate to
hypochlorite in the aged stock.

[ClO_{3}^{-}]/[OCl-]
= 0.364M/1.68M = 0.2161 M/M

Now it’s a simple matter of converting the 2.5 mg/L dose of chorine
to moles per liter (remember that chlorine is expressed as Cl_{2} and
thus has a GFW of 71) and then convert to mg/L of ClO_{3} (GFW = 83.45)

[ClO_{3}^{-}]
= __0.635 mg/L__

d. Repeat "c", but
assume that the chlorine stock was diluted by a factor of 4 prior to holding
for 40 days. Assume that dilution does
not change the value of the rate constant [we will find out later that it will
change a bit due to the change in ionic strength].

Again, you need to first calcuate the final hypochlorite
concentration in the stock

and

So at 40 days, using the lower initial stock concentration (2.776/4
= 0.694M), we get:

C = [OCl-] = 0.597
M

And recognize that the chlorate formed is equal to one-third times
the hypochlorite lost

[ClO_{3}^{-}]
= (Co-C)/3

[ClO_{3}^{-}]
= 0.03226 M

Now consider the concentration in the dosed water. For this you’ll want to calculate the molar
ratio of chlorate to hypochlorite in the aged stock.

[ClO_{3}^{-}]/[OCl-]
= 0.03226M/0.597M = 0.054 M/M

Now it’s a simple matter of converting the 2.5 mg/L dose of chorine
to moles per liter (remember that chlorine is expressed as Cl_{2} and
thus has a GFW of 71) and then convert to mg/L of ClO_{3} (GFW = 83.45)

[ClO_{3}^{-}]
= __0.159 mg/L__

e. Consider an MCLG (maximum
contaminant level goal) of 0.2 mg/L for chlorate. Determine the amount of dilution of the
chlorine stock, which will just allow compliance after 40 days holding time at
25°C.

Can be done analytically, or by trial and
error. For the latter, you can go back
to the original kinetic model, and re-calculate the concentrations based on
different dilution factors. In excell
this is quite easy if you had already set up the problem. The answer is:

__Dilution Factor = 3.2 __