Asked by Adry
                I don't get this problem...
6.0 N H2SO4 = ________ M
How can you solve for Molarity without mL?
Thanks, much appreciated!
            
        6.0 N H2SO4 = ________ M
How can you solve for Molarity without mL?
Thanks, much appreciated!
Answers
                    Answered by
            DrBob222
            
    Technically one can't solve for normality without knowing the reaction that is involved; however, we do it all the time anyway, with a few rules we follow.
For acids, N = equivalents/L where equivalents = molar mass/the number of H atoms and we assume ALL of the H atoms react. The long way around is this and I'll shorten it later.
M = moles/L so a 1M soln H2SO4 = 1 mol/L = 98g/1L
N = equivalents/L so a 1N soln H2SO4 = 1 equivalent = 98/2/1L or 49 g/L
So you see 98 g gives us a 1M soln and it gives us a 2N soln. To convert then from one to the other
1M = 2N so M =N x #H atoms
or N = M/#H atoms. This is based on ALL if the hydrogen atoms reacting. If you wonder if H2SO4 can react with only 1 H atom, the answer is yes. We can replace only one H and if we do that the normality of the solution changes (while the molarity does not). Also that is why, technically, we can't assign a normality to a solution without knowing the reaction.
    
For acids, N = equivalents/L where equivalents = molar mass/the number of H atoms and we assume ALL of the H atoms react. The long way around is this and I'll shorten it later.
M = moles/L so a 1M soln H2SO4 = 1 mol/L = 98g/1L
N = equivalents/L so a 1N soln H2SO4 = 1 equivalent = 98/2/1L or 49 g/L
So you see 98 g gives us a 1M soln and it gives us a 2N soln. To convert then from one to the other
1M = 2N so M =N x #H atoms
or N = M/#H atoms. This is based on ALL if the hydrogen atoms reacting. If you wonder if H2SO4 can react with only 1 H atom, the answer is yes. We can replace only one H and if we do that the normality of the solution changes (while the molarity does not). Also that is why, technically, we can't assign a normality to a solution without knowing the reaction.
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