GChem: Normality question

[mprs]

---

Members don't see this ad.

Ran into this question in one of my practice tests.

What is the Normality and molarity of a 0.1L H2SO4 solution that requires 0.075L of 0.5 N KOH for complete neutralization.

------------

At first I tried to figure out the number of moles involved. So from the KOH,

0.075*0.5 = gram equivalent.

However, not quite sure how this can relate to the neutralization. Is that the number of moles of neutralization, or is it half the moles, or double?

Thanks in advance!

 

[UndergradGuy7]

Ran into this question in one of my practice tests.

What is the Normality and molarity of a 0.1L H2SO4 solution that requires 0.075L of 0.5 N KOH for complete neutralization.

------------

At first I tried to figure out the number of moles involved. So from the KOH,

0.075*0.5 = gram equivalent.

However, not quite sure how this can relate to the neutralization. Is that the number of moles of neutralization, or is it half the moles, or double?

Thanks in advance!

(0.075L KOH)(0.5N KOH)=(x N of H2SO4)(0.1L)
solve for x get the normality of H2SO4.

Then molarity*equivalents=normality

so N/2= molarity


If you want to do it the stoichiometry way. Then write the equation...

H2SO4 + 2KOH -> 2 H2O + SO4^2-

So if 0.075N KOH * (0.5N KOH) = moles of KOH
Then moles of KOH*(1mol H2SO4/2moles KOH)=moles of H2SO4

So the molarity of H2SO4 = moles of H2SO4/0.1L
and the normality = molarity*2

 

[mprs]

Okay perfect thanks. I figured out what my misconceptions is. In terms of normality, N1 V1 = N2 V2 will ALWAYS be true, which is not the case for morality.