I want to know how I would prep 500ml of 25% weight/volume potassium hydroxide. From this stock, how would I perpare a 0.1M working solution?
If you could give as much detail on working out too please. I'd like to understand.
A 25% (w/v) solution means 25g of solute in each 100ml.
Preparing 500 ml of 25% (w/v) KOH requires 25*5= 125 g of KOH (five times the amount to make 100 ml. This weight has to be dissolved in about 500 ml of water. That could be done by first solving it in about 400 ml of water and then diluted to get a final volume of 500ml. (Take also in consideration the purity of the reagent... For ex: If your reagent is 95% pure you will need 125/0.95= 131.59g of KOH 95% purity.). Then you will have your stock solution.
Now for the working solution:
A solution of 0.1 M KOH contains 0.1 mol KOH each 1000ml.
Molecular weight of KOH is 56.1 g.
So, 0.1 M KOH contains 5.61 g NaOH each 1000ml.
To get 1000ml of KOH 0.1 M you need to take 22,44 ml of stock solution... as:
0.1 M KOH * (56.1 g KOH / 1 mol) * (1000 ml KOH 25% / 250 g KOH) = 22,44 ml KOH 25%
Perhaps of some interest for your query:
https://www.researchgate.net/post/How_to_Prepare_1_M_KOH_solution_with_85_KOH_pellets
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