KI Calculation help

Psymon · Mar 18, 2019

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.

AllSignsPointToFish · Mar 19, 2019

I calculated that you will need 30,000 ug of iodide to raise the level in a 500L tank from 0 ug/L to 60 ug/L (60ug/L x 500L).

1g KI dissolved in 1L water should yield 0.7645g I/L solution, or 7.65x10^5 ug/L.

30000 ug/7.65x10^5 ug/L equals 0.03924L or 39.24 mL to your solution to raise the iodide concentration 60ug/L.

Looks like your math is correct unless we both did it wrong!

AllSignsPointToFish · Mar 19, 2019

The truth is that we neglected solubility effects which should be small since this is a dilute solution, and the fact that the potassium iodide is only 99% pure should actually increase the required volume slightly.

I think I can still say with confidence that somewhere between 39 and 40 mL are required to high the targeted concentration of 60 ug/L

Psymon · Mar 20, 2019

Hey. That's awesome. Thank you for checking that AllSignsPointToFish.

If the above is true, could you say the same is true for these two then? (Same formula used)

Zn Formula for Zinc ZnCl2
Setpoint 4 ug/l
48.2 Zn % by weight in ZnCl2
1 g of ZnCl2
= 482 ppm of pure
Solution 1 0.482 ppm of Zn diluted in 1l of H20
= 482 ug/l in 1ml of solution
0.008298755187 Needed Solution Strength
4.149377593 ml to raise Zn to Setpoint ug/l for a 500l tank

Mn Formula for MnCl2.4H20
Setpoint 2 ug/l
28 Mn % by weight in MnCl2.4H20
1 g of MnCl2.4H20
= 280 ppm of pure
Solution 1 0.28 ppm of Mn diluted in 1l of H20
= 280 ug/l in 1ml of solution
0.007142857143 Needed Solution Strength
3.571428571 ml to raise Mn to Setpoint ug/l for a 500l tank

AllSignsPointToFish · Mar 20, 2019

I calculated approximately 4.2 mL of the zinc solution you described above will be required.

I also calculated approximately 3.6 mL of the manganese solution will be needed.