A fellow sea traveler is struggling with an unsolved problem; I’d like to ask everyone for advice
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Prepare 50 g of an alkaline mixture with a content of 25% based on Na2O, such that the ratio of NaOH to KOH is 1:5. How many grams of NaOH with 30% purity and how many grams of KOH with 90% purity are required?
I would like to ask whether, when referring to an alkaline solution on a Na2O basis, KOH is considered equivalent to NaOH? If that is the case, then 0.24 moles of Na2O are required, which corresponds to 0.48 moles of OH-. Accordingly, 0.08 moles of NaOH and 0.4 moles of KOH are needed; these amounts can be converted to 10.67 units of 30% NaOH and 24.89 units of 90% KOH.