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Education/Training + Get AlertsWhen operators are faced with licensing exams, they sometimes struggle with test anxiety because they don’t fully understand how to properly study and prepare for exam questions. Here we take a look at a sample water treatment licensing exam question, how to solve the problem, and the correct answer.
Sample Question:
You are dosing fluoride in a finished water supply using sodium fluoride, which is 97 percent pure. The treatment rate is 32 mgd and the natural fluoride in the untreated water is 0.4 mg/L. If you aim to have a finished water concentration of 0.9 mg/L, what is your dose? The fluoride ion has an equivalent weight of 19 and sodium is 23 equivalent units.
a) 133.4 lbs/day
b) 137.6 lbs/day
c) 296.4 lbs/day
d) 305.6 lbs/day
When dosing fluoride, it is critical that the fluoride ion is considered and not the compound of the chemical being used. It is also important that the dose be very accurate. In this particular problem, your goal is a total 0.9 mg/L but 0.4 mg/L is already present.
So first, we need to establish the difference between our concentration goal and what is already present.
0.9 mg/L - 0.4 mg/L = 0.5 mg/L
Thus 0.5 mg/L is the dose.
Now we can apply this information to our standard dosing formula:
0.5 mg/L X 32 mgd X 8.34 lbs/gal = 133.4 lbs
But, we need to dose the fluoride ion not the chemical:
(Fl) 19 ew + (Na) 23 ew = 42 ew total, 19/42 = 0.45
This represents the fluoride ion concentration in the sodium fluoride chemical.
Thus, 133.4 lbs/0.45 = 296.4 lbs/day, but all fluoride compounds used in water treatment are not 100 percent pure. Inert chemicals are added to preserve and stabilize the compound. In this case, the compound is 97 percent pure so:
296.4 lbs/0.97 = 305.6 lbs/day.
Therefore the correct answer is d) 305.6 lbs/day.
About the Author

Mike Smith is program coordinator and lead faculty of the Water Quality Management program at Red Rocks Community College in Lakewood, Colo.
He has been in charge of the multi-faceted training program since 1996.