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Normality headache

I do not fully grasp the idea of Normality.

What are the normality and the molarity of a sulfuric acid solution if 30mL of it are required to completely neutralize 24mL of 0.02N NaOH? (one of them, N or M, is never equation dependent because...??)

HCl(aq) + NaOH(aq)--> NaCl(s)+H20(L)


You don't have to understand Normality to work these, in fact, many texts are not including it. It is a handy concept. Normalality depends on the concept of "hydrogen equivalents", that is the number of proton equivalents the acid can donate (or base can accept).

Hydrogen equivalents Compound
1 HCl
1 HNO3
2 H2SO4
2 Ba(OH)2
2 Mg(OH)2
1 NaOH
3 Al(OH)3

Normality= Molarity*hydrogen equivalents.
So a 2 M solution of H2SO4 is 4N concentration (4N=2M*2)

The titration equation is
NormalityAcid*Volume = NormalityBase*Volume

NormalityAcid*30ml=.02N*24
Normality Acid= .02*24/30

MolarityAcid= Normality/HydrogenEquivalents, but the HEq of sulfuric acid is 2, fo
Molarity= abovenormality/2



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To add to the excellent information from Bob Pursley, consider the following:
Just as molarity = # mols/L, then
normality = # equivalents/L.
So how do we get # equivalents.

Just as # mols = grams/molar mass, then
# equivalents = grams/equivalent weight(mass).

The equivalent weight of a compound is the molecular weight (molar mass)/the number of replaceable hydrogen's.

Therefore, equivalent weights are as follows:
for HCl = molar mass/1;i.e., molar mass and equivalent mass is the same.
H2SO4 = molar mass/2 if both hydrogens are replaced. so the molar mass is 98 and the equivalent mass is 49.
H3PO4= molar mass/3 if all three hydrogens are replaced. So the molar mass is 98 and the equivalent mass is 98/3. One note here. You note I continue to reference "replaceable" hydrogens. So it is possible for H3PO4 to have three different equivalent masses. If I replace all three hydrogens as in this titration,
3NaOH + H3PO4 ==> Na3PO4 + 3H2O
then the equivalent mass is molar mass H3PO4/3 = 98/3.

If I replace only two, as in this titration,
2NaOH + H3PO4 ==> Na2HPO4 + 2H2O
then the equivalent mass is molar mass H3PO4/2 = 98/2.

If I replace only one, as in this titration,
NaOH + H3PO4 ==> NaH2PO4 + H2O
then the equivalent mass is molar mass H3PO4/1 = 98/1. <i/>

I'm not sure about your statement in parentheses. Probably you are alluding to the mL x N = mL x N and if you know any three then the fourth can be calculated. That is what Bob Pursley meant when he said you didn't need to know anything about normality to work these type problems. Personally I am quite upset that texts are omitting this from their content for it is a useful concept. I am equally upset about text omitting the concept of gravimetric factor but that's a different story. Good luck and post any particular problems you have. Please continue to do what you have been doing recently; i.e., show your work. We can do a much better job of helping you if we see your work.




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