Solids disolved in liquids

  • #1
Hello all,
This is my first post here. My apologies if this is in the wrong place.
I'm building an electrolysis device using potassium hydroxide (KOH) as the electrolyte and I don't know the formula for determining the correct percentages.
I'm trying to dilute caustic potash from 45% KOH & 55% water to 28% & 72% respectively.
The weight is 1441.5 grams per liter (gpl).
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
Let's say that you start with 1L of 45%( that's by wiehgt, I'm guessing) solution and add x cc (=x gms) of water. You now have 450 g KOH with (550 + x) g H2O. So, the new dilution is 450/(550+x) = 28%

Solving for x gives, 0.28x = 450 - (0.28*550) = 296, or x = 1057.143 cc or 1.57 L

So, for every liter of solution, you must add 1.57L of water to get to the right dilution.
 
  • #3
Thanks you're a big help!
 
  • #4
Borek
Mentor
28,676
3,167
Try CASC, it does exactly such calculations.
 

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