High school Chemistry help (Equivalent masses/normality with acids/bases)

[SnoringFrog]

I don't really get this concept.

I understand how to get the eq mass. The eq mass of NaOH is ~40g right?

Ok, here's the questions I don't know if I answered right.

How many eq masses of NaOH would it take to completely neutralize 5 equivalent masses of H[SIZE="1"]2[/SIZE]SO[SIZE="1"]4[/SIZE]?

I thought 5 at first, or is it 10?

I also don't get normality or how it relates to molarity.

For molarity, normality = the number of equivalents per mole? So H[SIZE="1"]2[/SIZE]SO[SIZE="1"]4[/SIZE] would be 2, HCl would be 1, and something like H[SIZE="1"]3[/SIZE]PO[SIZE="1"]4[/SIZE] would be 3? And how does that work with bases? Is NaOH 1 or does that not work?

My books says normality = equivalents of solute/liters of solution. So the normality of 5 liters of H[SIZE="1"]2[/SIZE]SO[SIZE="1"]4[/SIZE] would be 2/5 or .4, correct?

I'm sorry if any of this doesn't make sense, I really don't get chem very well. And thanks in advance for any help you can give!

[Whitefang]

It has been a long time since I studied chemistry, so please do not take this as gospel. Also, remember to work out the problem yourself and learn to apply it generally or I haven't really helped you here. If you need me to make this clearer, let me know.

Well, just by looking at it, I can guess that one of the products of a reaction is H2O. After googling the other product is Na2SO4. (I leave this to you to figure out why, as it is vital you know why this occurs so that you can apply this to any equation)

So, if we take H2SO4 + NaOH = H2O + Na2SO4, all we have to do is balance for equilibrium.

We first take Sodium, and see that we need twice as many NaOH molecules as resultant Na2SO4 molecules (Twice as many Na Sodium atoms). By the same principle, we need only one H2SO4 molecule for every Na2SO4 molecule produced. This leaves us with 2 O atoms and 4 H atoms left, which means that we balance the equation as such.
H2SO4 + 2 NaOH --> 2 H2O + Na2SO4

So if we need twice as much NaOH for a reaction to occur, then it follows that you need 10 equivalent masses if you are to neutralize 5 H2SO4.

[SnoringFrog]

I understand what you're saying there, but the only problem with that is that I won't be able to figure out the products of the reactions...the question I posted is exactly in the form I'm expecting the test to have.

So is there a way to know w/o having a full equation to balance? Looking at it I think I see the pattern. NaOH has one ionizable atom of OH, and H2SO4 has two of H. Since H2SO4 has twice as many, NaOH will need twice as many equivalent masses. Is that thinking correct? Because it also leads to an answer of 10.

[Whitefang]

Yes, I believe that works. First, though, remember that OH- is not an atom! I'm sure you knew that, but it can cost you points sometimes depending on the teacher.

You should be able to break the molecules apart into ionized sections. Na+ and OH-, and 2 H+ and SO4(2-), will lead you to see that you need 2 NaOH for 1 H2SO4. (I wish I knew HTML subscript and superscript, it would make this easier to see.) You should definitely study common molecules that are made up of ions. It could help you out.

[SnoringFrog]

Oh, right! It's a molecule. I knew that. lol

Eh...we kinda studied that back at the beginning of the year, but I've forgotten most of those ionic compounds now (which is killing me in this chapter) and my test it tomorrow...this is a bit late to realize I don't know this, but I'm trying to get what I can in time.

What about the normality stuff?

I also don't get normality or how it relates to molarity.

For molarity, normality = the number of equivalents per mole? So H2SO4 would be 2, HCl would be 1, and something like H3PO4 would be 3? And how does that work with bases? Is NaOH 1 or does that not work?

My books says normality = equivalents of solute/liters of solution. So the normality of 5 liters of H2SO4 would be 2/5 or .4, correct?

[Whitefang]

I'm afraid I don't remember anything about normality, sorry. Like I said, it's been a while. Hopefully someone else can help you out.