posted by maddy on .
EXTRA CREDIT QUESTION(this is like kinda really hard for me at the moment i got the first one but the others not so much)
The city of Celina was ordered by the Ohio EPA in 2003 to resolve its MCL (maximum contaminant level) violations of its drinking water. The Celina water treatment plant mainly uses chlorine (Cl2) to help disinfect the water taken from Grand Lake. TTHMs or trihalomethanes are a disinfection byproduct that results when drinking water is chlorinated. The MCL for TTHMs in public drinking water is 80. ppb. During 2007 the TTHM concentration was as high as 235 ppb and the running average for the October 1 through December 31, 2008 monitoring period was 112 ppb . The newly constructed water treatment facility utilizing GAC (granular activated carbon) became operational in 2009 and is reportedly keeping the TTHM concentration below the MCL.
Some people who drink water containing trihalomethanes in excess of the MCL over many years may experience problems with their liver, kidneys, or central nervous systems, and may have an increased risk of getting cancer.
(b) When Celina's water was running an average of 235 ppb, how many milligrams per liter was this over the limit?
(c) If Celina lowered its TTHMs down to 54.0 ppb, what volume of water would be required to equal the 0.080 mg/L of the MCL? (Hint: use the dilution formula to solve this problem.)
(d) If the Celina water treatment plant produced 1300000 gallons of water over a 24.0 hr period, what total mass of TTHMs would this water contain if the TTHM level was at 50.0ppb? (1.00 gallon = 3.785 liters)
(e) If the level of TTHMs in the water is 99 ppb, and a GAC cartridge is capable of removing 25 Kg of TTHMs, what volume of water can be reduced to a TTHM concentration of 70. ppb by one cartridge?
(f) If the cartridge in the previous question processes 4,000,000 liters of water per day, how many days can it operate before it reaches it's filtering capacity?
b) 235 ppb = 0.235 mg/L
How much is that over 0.080 mg/L?
c) 0.054 x ?L = 0.080. Solve for ?L.
d) 1.3E6 gallons x 3.785 L/gallon x 0.050 mg/L = ?
You're on your own for the others. This should give you some clues on how to attack the others. Note that dimensional analysis works wonders.