Charles Harding
- These are extremely dangerous solutions you're making. They can cause serious burns and eye damage; wear gloves and goggles, OK?
- You'll also find that these compounds generate a LOT of heat when they get wet- so much so that they can boil over, spraying NaOH or KOH everywhere. Lovely. Keep the solution cold as you add the water.
- As soon as you pour the NaOH or KOH into a beaker, it will start to suck CO2 and water from the air. That will change its mass and introduce a lot of uncertainty into the calculation I'm going to outline for you. Don't expect any more than two significant figures in the concentration. You'll have to do a titration if you want more precision than that.
The normality (N) of the solution is the number of equivalents dissolved in one liter of solution. Your solutions involve strong bases, so in this case "equivalents" means "moles of hydrogen ions that can be consumed". One mole of hydroxide ions (OH-) can react with one mole of hydrogen ions, so you can write "1 equivalent of NaOH = 1 mol of NaOH".
Here's one way to solve the problem.
- What's the unknown? You want to know how many grams of sodium hydroxide to weigh out to make your first solution.
- What's given? The concentration is 5.0 N NaOH. There's something missing, though.
The amount of sodium hydroxide you'll need is obviously going to depend on how much solution you want to make up. I'll assume you want 1 L:
1 L solution ? 
g NaOH - List relationships that connect the given information with the unknown. Start from the unknown. You can relate
g NaOH to moles of NaOH with the molecular weight of NaOH:
1 L solution ? mol NaOH molecular weight
of NaOHg NaOH 1 L soln 1 L soln = 5 equiv NaOH equiv. NaOH 1 equiv. NaOH = 1 mol NaOH mol NaOH molecular weight
of NaOHg NaOH - Do the math. Can you take it from here? Set up each of the links in the roadmap as conversion factors. All of the units should cancel to give you g NaOH in the end.
- Check the answer. Is it reasonable?