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I can get 80% pure sodium chlorite and I want to make a 28% solution of it. Though I'm getting some conflicting information on the web, this site seems to indicate that 28 grams of pure NaClO2 to 72 grams of water would give me what I want, and since a gram of pure water is 1ml at STP, I could measure the water as volume rather than weight.

But since I have 80% NaClO2 powder, does that mean I can just multiply the amount by 1/.8 = 1.25, adding 35 grams of powder to 72 ml of water, to get a 28% nominal solution?

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I'm assuming your Sodium Chlorite is 80% pure by mass, and you want your solution to be 28% pure by mass as well.

Let's say you have 1 g of powder and let's calculate how much water you need to add to make a 28% solution (you can then proportionally scale up both amounts.)

Let $y$ be the number of grams of water added. Then the total mass is $1+y$. The mass of $\textrm{NaCLO}_2$ stays constant at .8. So we want to solve $$\frac{0.8}{1+y} = 0.28$$ which has solution $y \approx 1.86$ grams.

So if you have e.g. 35 g of powder, you will need to add about $65.1$ ml of water to make a 28% solution.

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OK, so the other impurities are treated the same as the primary impurity, water, in that regard. makes sense, thanks! –  jcomeau_ictx Nov 6 '11 at 23:35

I add this link. Maybe it is useful to clear out the calculation of the quantities to get a $28\%$ solution of NaCLO2. In my understanding a $80\%$ sodium chlorite powder is dissolved in distilled water. The quantities are: $28\%$ sodium chlorite powder ($80\%$) by weight and rest of $72\%$ distilled water by weight.

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