Let's do the arithmetic, TrashKan.
Let's say 1 T = 15 g, so 2 T = 30 g, Let's further agree that 1 C = 230 ml = 230 g , and so 1/2 C = 115 g. Let's agree that 15g of Bentonite + 115 g of water = 130 g of solution, and so if 1 t = 5 g which is 5/130 of the solution. = 0.038 of the solution you are asked to apply to each gallon. To apply this then to FOUR gallons, you need to multiply 0.038 by 4 = or .153 9or more than 1/10th of the volume of the solution Bentonite asks you to make.
But , you mixed 9T which is 135 g in 2 C (which = 460 g) and so the total volume/weight = 595 g total, 1 T = 15 g which is 15/595 = 0.025 which is what you want to apply to 4 gallons. You need to add SIX times the 1 T , or 6 T to add the correct amount of the clay slurry... That's my story and I am sticking to it.