Molarity of solution?

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

khadcheemz

Full Member
7+ Year Member
Joined
Jul 8, 2015
Messages
49
Reaction score
10
What is the molarity of a solution of H2SO4 if 100 ml of this solution requires 25 mL of 2 M NaOH for complete neutralization?

Isn't this N1V1=N2V2?

I thought it was 1.0 M but its .25 M

Members don't see this ad.
 
(2mol/L of OH-)(25mL of solution) = 0.5mmol of OH = 0.5mmol of H+
(0.5mmolH+) (1mmol H2SO4/2mmolH+) = 0.25mmol H2SO4
0.25mmolH2SO4/100mL = 0.25M of H2SO4 initial.
 
You can do it this way because NaOH is a strong base, so you can assume it dissociates 100%.
H2SO4 is a strong acid you have to assume the it also dissociates 100% twice.
 
You can do it this way because NaOH is a strong base, so you can assume it dissociates 100%.
H2SO4 is a strong acid you have to assume the it also dissociates 100% twice.
You still do 2 N but I'm not sure if it dissociates 100% the second time.
 
Members don't see this ad :)
Yeah, in the real world you know the second H doesn't dissociate 100% but i guess in terms of the DAT you would have to assume it does.
Oh ok just checking thanks. On a pH problem they could potentially ask it like that right? Then we would treat the second H as not a complete [H+] right?
 
Oh ok just checking thanks. On a pH problem they could potentially ask it like that right? Then we would treat the second H as not a complete [H+] right?
That's a really good question. I think if its a strong acid, you can assume this. If it isn't a strong acid you can't assume this.
If you have a specific question, I think it would be best to ask Dr. Romano.
Also, if they don't give you any pKa's or Ka's, your going to have to assume it dissociates twice.
 
If you have DAT destroyer, I'm sure there are a few questions like this.
I do, but I don't recall a question involving the pH of a diprotic other than like Ba(OH)2. But I think they treated it as a total OH. Should be the same type of approach though right?
 
Top