Sciencemadness Discussion Board

Noob Solution Percentage Question

arkallic - 30-1-2014 at 16:00

Hey everyone.

So this is kind of a noob question.

I'm gonna be doing a project that calls for "10ml of a 10% solution of KOH in 95% ethanol"

So here's my question. What's the right way to figure out how much KOH to put in the 10ml of ethanol?

If the ethanol is 7.89g, do I just take 10% of that (.79g)?

Or do I go by moles?

I did the math in moles and got about .94g of KOH.

So what way is right? Or with such a small difference does it even matter?

[Edited on 31-1-2014 by arkallic]

DraconicAcid - 30-1-2014 at 16:11

The 10% is almost certainly by mass (although some chemists would use mass/volume percent, which is easy, yet ridiculous), never moles. That would give you 0.8 g of KOH, or 1 g KOH if it's mass/vol%. In most applications, it probably won't matter, because your potassium hydroxide isn't pure anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)

arkallic - 30-1-2014 at 16:15

Thanks very much!

blogfast25 - 31-1-2014 at 14:05

Quote: Originally posted by DraconicAcid  
, because your potassium hydroxide isn't pure anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)


Actually, commercial KOH almost always contains 10 w% water.