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Solids disolved in liquids

  1. Oct 5, 2005 #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).
  2. jcsd
  3. Oct 5, 2005 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  4. Oct 5, 2005 #3
    Thanks you're a big help!
  5. Oct 5, 2005 #4


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    Staff: Mentor

    Try CASC, it does exactly such calculations.
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