So from my calculations, an area of 80' by 20' (1600 sq ft) at 4' deep would need 1.28lbs of Aquathol Super K. This can be calculated as (application rate in ppm) x (average depth in feet) x (0.16). The .16 is for 1600 square feet. The PPM is 2.0.

2.0 x 4.0 x 0.16 = 1.28 pounds for 1600 sq feet at 4' deep

That would give me about 7.8 times from a 10lb bag of Super K.

The Aquathol K liquid on the other hand offers significantly more value and would require about 24oz per 1600 sq feet at 4' deep. I calculated this based on the SDS label which shows the ppm rate at 3.8 oz per 1000sq feet.

So to calculate 1600 sq ft I used the following:

(3.8 / 1000) * 1600 = 6.08 oz per square foot. Then multiply this by the depth of 4'. This totals 24.32 oz at a 4ft depth for 1600 sq ft.

A 2.5 Gallon jug would be 320 oz which would allow me to treat more than 13 times. The liquid is also less costly so I figure it would be nearly half the price to apply the liquid as opposed to the crystals.

Unfortunately, I have no boat to sit in and inject the liquid below the surface as you suggested. I suppose I could get a canoe but I don't really have a place to store a boat. I could get some waders and wad in with a backpack sprayer. Otherwise, it seems like the crystals may be my best option?

One of my friends has a lake the same size as mine (5 acres) and he told me that once a year in May, he pours 2.5 gallons of Aquathol (half on each side) into his pond and treats the rest with 2.5 gallons of Tribune (I think that's just Diquat) and roundup. He told me that's enough to treat all of the submerged weeds in the lake. I thought that seemed like way to little to do anything and was the amount for maybe 1/2 acre.