CEE 680 16 November 1999

SECOND EXAM

Closed book, two pages of notes allowed.

Problem #1: (40%) Carbonate System.

Two raw drinking waters are mixed in equal proportions as they enter the headworks of a water treatment plant. The two are characterized as follows:

 Water Alkalinity (mg/L as CaCO3) pH A 70 7.31 B 220 9.12

What will the pH of the blended water be?

Solution to Problem #1

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 7.31 0.9101 0.00093 9.12 0.9367 0.0619

so, for water A:

CT = {(70/50,000) - (10-6. 69) + (10-7..31)}/(0.9101 + 2*0.00093) = 0.001535 M

and for water B:

CT = {(220/50,000) - (10-4.88) + (10-9.12)}/(0. 9367 + 2*0.0619) = 0.004161 M

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

Alk = (70 + 220)/2/50,000 = 0.00290 equ/L

CT = (0.001535 + 0.004161)/2 = 0.002848 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 0.00290–0.002848 = 0.000052

and (Alk – 2CT) is 0.00290–2(0.002848) = –0.002796

or

pH = 8.66

Problem #2: (60%) Precipitation/Dissolution

Aluminum Solubility

1. Prepare a solubility diagram for Aluminum Hydroxide (amorphous) in water. Present it in the usual form (log C vs pH). Outline the zone of precipitation.
2. Prepare a solubility diagram as above in Part 1, but this time assume that 0.1 mM total phosphate is present. Outline the zones of precipitation and mark the identity of the precipitates.

 Equilibria Log K AlOH3(s) (amorphous) + 3H+ = Al+3 + 3 H2O 10.8 Al+3 + H2O = AlOH+2 + H+ -4.97 Al+3 + 2H2O = Al(OH)2+ + 2H+ -9.3 Al+3 + 3H2O = Al(OH)3(aq) + 3H+ -15.0 Al+3 + 4H2O = Al(OH)4- + 4H+ -23.0 AlPO4•2H2O(s) (variscite) = Al+3 + PO4-3 + 2H2O -21 H3PO4 = H+ + H2PO4- -2.15 H2PO4- = H+ + HPO4-2 -7.2 HPO4-2 = H+ + PO4-3 -12.3

Solution to Problem #2

The above equilibria can be re-arranged to get a series of equations expressing the log concentration of each species as a function of pH.

First do this for the species in equilibrium with amorphous aluminum hydroxide.

 Components Log K Species Al(OH)3 (S) H+ PO4 2 Al 2 1 3 0 10.8 Al(OH) 1 2 0 5.83 Al(OH)2 1 1 0 1.5 Al(OH)3 1 0 0 -4.2 Al(OH)4 1 -1 0 -12.2 H+ 0 1 0 OH- 0 -1 -14

Which gives rise to the following lines:

 Log[ Al 2 ] = + 10.8 -3 pH Log[ Al(OH) ] = + 5.83 -2 pH Log[ Al(OH)2 ] = + 1.5 -1 pH Log[ Al(OH)3 ] = - 4.2 0 pH Log[ Al(OH)4 ] = - 12.2 +1 pH Log[ H+ ] = 0 -1 pH Log[ OH- ] = - 14 +1 pH

Next do this for the species in equilibrium with aluminum phosphate.

 Components Log K Species AlPO4 H+ PO4 2 Al 1 0 -1 -21 Al(OH) 1 -1 -1 -25.7 Al(OH)2 1 -2 -1 -30.3 Al(OH)3 1 -3 -1 -36 Al(OH)4 1 -4 -1 -44 H+ 0 1 0 OH- 0 -1 -14

Which gives rise to the following lines:

 Log[ Al ] = -21 +0 pH -1 Log PO4 Log[ Al(OH) ] = -25.7 +1 pH -1 Log PO4 Log[ Al(OH)2 ] = -30.3 +2 pH -1 Log PO4 Log[ Al(OH)3 ] = -36 +3 pH -1 Log PO4 Log[ Al(OH)4 ] = -44 +4 pH -1 Log PO4 Log[ H+ ] = 0 -1 pH 0 Log PO4 Log[ OH- ] = -14 +1 pH 0 Log PO4