Normality/Molarity question from Achiever

[YourDentition]

Hi, This a question from Achiever that seems to give an answer different from what i previous knew about difference between Normality and Molarity.
I would appreciate if someone would explain why the following is the answer to this question. Thank you in advance.


Determine the normality and molarity of a 100.00 ml H2SO4 solution that requires 75.00 ml of 0.5000 N KOH for complete neutralization.


Achiever's answer and explanation:
Normality: V1N1 = V2N2
(75*.5)/100 = .375N

Molarity:
.375/2 = .18M

Now since each mole of diprotic H2SO4 is associated with 2 equivalents of acid, why did they divide by 100ml instead of 200ml to get our normality??

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[nze82]

Hi, This a question from Achiever and seems to give an answer that contradicts my previous knowledge about the difference between Normality and Molarity.
I would appreciate if someone would explain why the following is the answer to this question. Thank you in advance.


Determine the normality and molarity of a 100.00 ml H2SO4 solution that requires 75.00 ml of 0.5000 N KOH for complete neutralization.


Achiever's answer and explanation:
Normality: V1N1 = V2N2
(75*.5)/100 = .375N

Molarity:
.375/2 = .18M

Now since each mole of diprotic H2SO4 is associated with 2 equivalents of acid, why did we not divide by 200 instead of a 100ml to get our normality??

N >= M

In this case:

N = 2M

So:

M = N/2

 

[ethanyoo726]

Use this equation to find out Normality. N1 is what we are trying to find out.

N1V1 = N2V2

(N1) * (100) = (.5) * (75)
-----> N1 = 3/8 = 0.37 N

Now we know that H2SO4 is 2N due to the 2 Hs that can dissociate. So N1 = 2* M. So the Molarity (M) is just 0.37/2 = .18 M

hope that helps.