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I couldn't find a Stack Exchange site for computational chemistry, so hope this is apropos for this forum.

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 $28\,\rm g$ of pure $\mathrm{NaClO_2}$ to $72\,\rm g$ of water would give me what I want, and since a gram of pure water is $1\,\rm mL$ at STP, I could measure the water as volume rather than weight.

But, since I have $80\%$ $\mathrm{NaClO_2}$ powder, does that mean I can just multiply the amount by $1/.8 = 1.25$, adding $35\,\rm g$ of powder to $72\,\rm mL$ of water, to get a $28\%$ nominal solution?

andselisk
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jcomeau_ictx
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2 Answers2

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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 $1\,\rm g$ 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 $\mathrm{NaClO}_2$ stays constant at $.8$. So, we want to solve $$\frac{0.8}{1+y} = 0.28$$ which has solution $y \approx 1.86\,\rm g$.

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

andselisk
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user7530
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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 06 '11 at 23:35
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I add this link. Maybe it is useful to clear out the calculation of the quantities to get a $28\%$ solution of $\mathrm{NaClO_2}$.

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.

andselisk
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user57116
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