How to Determine Correct Percentages for Diluting Potash in Electrolysis Device?

  • Thread starter Thread starter dkmacmillan
  • Start date Start date
  • Tags Tags
    Liquids Solids
Click For Summary

Discussion Overview

The discussion revolves around determining the correct percentages for diluting potassium hydroxide (KOH) in an electrolysis device. Participants explore the calculations necessary to achieve a specific dilution from a concentrated solution.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant describes their initial setup, indicating a desire to dilute a 45% KOH solution to 28% and seeks guidance on the calculation method.
  • Another participant proposes a mathematical approach to find the amount of water needed for dilution, providing a formula based on the initial and desired concentrations.
  • The calculation involves starting with 1L of a 45% solution and determining the volume of water to add to achieve the target dilution.
  • A suggestion is made to use a specific tool, CASC, which is said to perform such calculations.

Areas of Agreement / Disagreement

Participants do not express any disagreement on the mathematical approach presented, but the discussion does not reach a consensus on the best method or tool for performing the calculations.

Contextual Notes

The discussion does not clarify any assumptions regarding the density of the solutions or the specific conditions under which the calculations are valid.

Who May Find This Useful

Individuals interested in electrolysis, chemical dilution processes, or those seeking to understand the calculations involved in preparing electrolyte solutions.

dkmacmillan
Messages
3
Reaction score
0
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).
 
Chemistry news on Phys.org
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.
 
Thanks you're a big help!
 
Try CASC, it does exactly such calculations.
 

Similar threads

How does electrolysis affect salts in solution?
  • · Replies 3 ·
Replies
3
Views
3K