 # 1. Determine the gallons/day of alum solution (Al2(SO2)3 18H20) - (...

## Question

1. Determine the gallons/day of alum solution (Al2(SO2)3 18H20) - (density of 80 lb/ft3) required to precipitate the phosphorus in wastewater which contains 10 mg P/L. Based on lab tests, 2.5 moles of Al are needed per mole of P. The alum strength is 50% and the flow rate of the wastewater stream is 3.5 x 106 gallons/day.
2. For question 1 above, calculate the required storage capacity if a 30-day supply of alum solution (Al2(SO2)3 * 18H20) is to be stored at the treatment facility
3. Verify by calculations that the 16 g of sodium phosphate* 12H2O you added to 80 L of water in your experiment produced a phosphate ion concentration of about 50 mg/L.

## Solution Preview

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There are many parameters that need to be considered in this exercise, so it is useful to be consistent and convert units as we go to make the calculation convenient.
Let’s begin with the concentration of phosphorus (P) given. We have information that the concentration of P in wastewater (WW) is 10 mg/ L:
C(P)=10 mg⁄ L
Also, there are 3.5 × 106 gallons of WW produced daily, or:
V(WW)=3.5 × 106 gallons / day=371 gallons ⁄ day
So, within 1 day we have 371 gallons of WW, which contains P with a concentration of 10 mg/L. We can calculate the mass of P in the daily amount of WW based on the above data. However, to do that, we need to convert gallons of WW to liters first:
V(WW)=371 gallons=1404 L
Now, knowing that there are 10 mg of P per each 1 L of WW, the mass of P that should be treated with alum daily, in g, may be calculated as follows...

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