hi R2R Chemistry and maths pros,
Please, can someone check my maths on this calc.
I have some KI (German:Kaliumiodid (min. 99%, Ph. Eur., USP) 100g)
Calculations are made using 100%.
If the Iodide setpoint is 60ug (Triton target)
Potassium Iodide = 76.4% iodide (source: https://www.convertunits.com/molarmass/KI)
1 g of KI
= 764 ppm Iodide
Solution 1: 1g of KI dissolved in 1L of H20 = 764 ppm of I
Therefore 1ml of Solution 1 = 0.764 ppm I or 764ug/l
formula 60(setpoint) divided by 764ug/l
=0.07853403141
Therefore
0.07853403141 x 500L (Volume in reef)
=
One would need 39.26701571 ml to raise Iodide from 0 to 60ug in 500L tank?
I hope that makes sense.
Please, can someone check my maths on this calc.
I have some KI (German:Kaliumiodid (min. 99%, Ph. Eur., USP) 100g)
Calculations are made using 100%.
If the Iodide setpoint is 60ug (Triton target)
Potassium Iodide = 76.4% iodide (source: https://www.convertunits.com/molarmass/KI)
1 g of KI
= 764 ppm Iodide
Solution 1: 1g of KI dissolved in 1L of H20 = 764 ppm of I
Therefore 1ml of Solution 1 = 0.764 ppm I or 764ug/l
formula 60(setpoint) divided by 764ug/l
=0.07853403141
Therefore
0.07853403141 x 500L (Volume in reef)
=
One would need 39.26701571 ml to raise Iodide from 0 to 60ug in 500L tank?
I hope that makes sense.