CEE 680 16 November 2000

# SECOND EXAM

Closed book, two pages of notes allowed.

# Problem #1: Carbonate System.

(50%) Two raw drinking waters are mixed as they enter the headworks of a water treatment plant.  The two are characterized as follows:

 Water Flow (MGD) Alkalinity (mg/L as CaCO3) pH #1 20 70 6.98 #2 10 220 8.85

A. What will the pH of the blended water be immediately after mixing?

B. What will the pH of the blended water be after it has reached equilibrium with the bulk atmosphere?

# Solution to Problem #1

## Part A.

This is a closed system problem.  Therefore the total carbonate concentrations (CT's) must be determined and treated as conservative.  Likewise the alkalinities are conservative, and then the final pH can be determined from the blended CT and alkalinity.

for either water:

first, I would determine the alpha's at the two pH's.  Recall the general equations for a diprotic acid are:

This results in the following values (assuming pKs of 6.3 and 10.3:

 pH alpha-1 alpha-2 6.98 0.8269 0.000396 8.85 0.9631 0.03417

so, for water #1:

CT = {(70/50,000) - (10-7.02) + (10-6.98)}/(0.8269 + 2*0.000396) = 0.001692 M

and for water #2:

CT = {(220/50,000) - (10-5.15) + (10-8.85)}/(0.9631 + 2*0.03417) = 0.004259 M

now we need to calculate the blended alkalinities and total carbonates:

Alk = (2*70 + 220)/3/50,000 = 0.00240 equ/L

CT = (2*0.001692 + 0.004259)/3 = 0.002547 M

Now calculate the pH from the earlier equation for total carbonates, and making the following simplifying assumption:

1.        Alk is large compared to [OH] and [H]

which becomes:

which is simplified to:

and (Alk – CT) is 0.00240–0.002547 = -0.000147

and (Alk – 2CT) is 0.00240–2(0.002547) = –0.002694

or

pH = 7.50

## Part B.

This is now an open system problem.

Charge balance considerations dictate that:

Alk = CT

Under conditions where pK1 << pH<< pK2, the following is approximately true:

CT = [HCO3-]

And from the equilibrium equation we can conclude:

[HCO3-] = K1 [H2CO3*]/[H+]

And since this is an open system, we can say:

[H2CO3*] = KH pCO2

which becomes, for the bulk atmosphere at 25oC

[H2CO3*] = 10-5

we can combine and get the equation for alkalinity (or bicarbonate) vs H+ in an open system

Alk = [HCO3-] = K1 10-5 /[H+]

Then substituting and isolating H+, we get

[H+] = 10-11.3/Alk

and substituting in for the blended water alkalinity

[H+] = 10-11.3/0.00240

[H+] = 2.088 x 10-9

or

pH = 8.68

# Problem #2: Complexation

1. (50%) You are trying to control algal growths in a lake containing water that is well buffered at neutrality (pH 7.00).  Accordingly, you have just added 10 µM of Cu+2 for this purpose.  However, you discover that the intended effect did not occur.  Careful analysis of this lake water showed a substantial presence of EDTA.  The total EDTA concentration (YT, the sum of all species) was measured as 20 µM.

A. How much free copper actually existed?  Assume the system is at equilibrium, and ignore any copper hydrolysis species.

B. How much manganese (+II) would have to be introduced into this water to reduce the degree of copper binding by EDTA to 50% of its total concentration (i.e., CuEDTA = 0.5*CuT)?

C. Describe in qualitiative terms how your approach to solving this problem would differ if the ligand was not EDTA but some monodentate ligand such as ammonia.  Lets presume that you do not have access to MINEQL or any similar computer code.

Important Equilibria

 Equilibrium Log K Y-4 + H+ = HY-3 10.26 Y-4 + 2H+ = H2Y-2 16.42 Y-4 + 3H+ = H3Y- 19.09 Y-4 + 4H+ = H4Y 21.09 Y-4 + 5H+ = H5Y+ 22.59 Y-4 + 6H+ = H6Y+2 22.59 Y-4 + Cu+2 = CuY-2 18.80 Y-4 + Mn+2 = MnY-2 13.87

# Solution to Problem #2

## Part A

The active chelating form of EDTA is the fully deprotonated species, Y-4, and its abundance is determined from the appropriate alpha value:

First translate the cumulative formation K’s (i.e., the beta’s) into the classical stepwise deprotonation K’s we use for acid/base calculations (i.e., the Ka’s)

 Equilibrium Log K Stepwise Deprotonation Equ. Log pKa’s Y-4 + H+ = HY-3 10.26 HY-3 = H+ + Y-4 -10.26 Y-4 + 2H+ = H2Y-2 16.42 H2Y-2 = H+ + HY-3 -6.16 Y-4 + 3H+ = H3Y- 19.09 H3Y-1 = H+ + H2Y-2 -2.67 Y-4 + 4H+ = H4Y 21.09 H4Y0 = H+ + H3Y-1 -2.0 Y-4 + 5H+ = H5Y+ 22.59 H5Y+1 = H+ + H4Y0 -1.5 Y-4 + 6H+ = H6Y+2 22.59 H6Y+2 = H+ + H5Y+1 0

Then use these values to determine the alpha

6  =  1/{[H ]6/K1K2K3K4K5K6 + [H+]5/K2K3K4K5K6 + [H+]4/K3K4K5K6 + [H+]3/K4K5K6 + [H+]2/K5K6 + [H+]/K6 + 1}

so for pH =7,

6  =  4.8 x 10-4

now consider the various equations that govern this problem:

Equilibrium equation:

Mass Balance equations:

where YTF is the total free Y, defined as we did in class for the NTATF.

A good way to approach a problem of this sort is to look for simplifying assumptions.  Mass balance, charge balance and proton balance equations are the best place to find these things.  I would presume based on the strength of EDTA complexes, and the fact that there is an EDTA excess, that nearly all of the metal will be bound.  In mathematical terms this is:

and then:

and now combining with the equilibrium equation:

which can be re-arranged to:

which for pH = 7, becomes:

## Part B

Now with this problem, we want half of the copper to be free, and half complexed with EDTA.  Since Mn also complexes with EDTA rather well, we should assume that any residual EDTA will be bound with the excess Mn.  This gives us the following mass balance approximations:

using the first two and returning to the CuY equilibrium, we can estimate the free Y-4.

Now we can use this along with the MnY equilibrium and the MnY assumption to get the free manganese

Since this is far greater than the MnY, it is essentially equal to the total manganese, and the answer to the problem is:

1.28 M of MnT is required (an excessively high amount).

## Part C

With a monodentate ligand you would need to consuder the possibility of multi-ligand metal complexes.  In this case, you would need to use an alpha diagram, and try to solve using the n-bar curves.