Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3Ca+2 + 3SO4-2 1. Some consumers might be concer

Question

Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3Ca+2 + 3SO4-2

1. Some consumers might be concerned about adding aluminum to their drinking water. If the pH of the finished water (following filtration, chlorination, and final pH adjustment) is 8.0, and if the solubility of Al+3 is governed by the solubility of Al(OH)3, how much dissolved Al+3 is expected to remain in solution? Assume pKs = 32.9 for freshly precipitated Al(OH)3(s).

2. How many pounds of alum will be used per day when the plant is operated at the design flow rate using an alum dosage of 30 mg/L? (Hint: there is a very useful conversion factor you can use to solve this problem, and future problems of this nature: (mg/L) x MGD x 8.34 = lbs/day. The basis for this conversion factor is that 1 mg/L = 1 ppm = 1 gal/MG and that 1 gal of water weighs 8.34 lbs. In making this calculation, be sure to express the flow rate in MGD, not gal/day, or your answer will be a million times too high. If in doubt, check your answer using routine conversion factors.)

3. The treatment plant has two identical circular sedimentation basins operated in parallel, with each one treating half of the flow (10 MGD). Each basin has a hydraulic retention time of 3 hours and is 18 feet deep. What is the diameter of the basins? (Hint: there are 7.48 gallons in 1 ft3.)

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st question for you.Please ask other as different question

1.
pH = 8
pOH = 14-8 =6

pOH= -log [OH-]
6 = -log [OH-]
[OH-]= 1*10^-6 M

Al(OH)3 ---> Al3+ + 3 OH-
                       
Ksp = [Al3+] [OH-]^3
32.9 = [Al3+] *(1*10^-6)^3
[Al3+] = 3.29*10^19 M
Answer: 3.29*10^19 M

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