From what you've said, I think the teacher has TOLD you that the HCl s/he has provided is 0.1046 N. And if you've used 10.15 mL (0.01015 L), the # equivalents of HCl is L x N = 0.1046 x 0.01015 = ?
Next, I assume you are to calculate the N of the NaOH listed next at 10 mL (0.01L).
Since we know that 1 equivalent of acid = 1 equivalent of base (actually = 1 equivalent of ANYTHING ELSE which is what make it nice when working with normality), then equivalents of the HCl above = equivalents of NaOH.
Since eq NaOH = L x N and we know equivalents NaOH and L NaOH, you can calculate the N or NaOH.
I obtained about 0.1062 N for NaOH.
Post a new question at the top of the page if I've misinterpreted your post.
I wanted to calculate the normality of the acid and base solutions for a lab. But I'm extremely confused. I know normality is number of equivalents divided by volume in liters. But are we suppose to be given the number of equivalents of HCl and NaOH? Or are we suppose to figure out the number of equivalents? My lab teacher said that HCl is 0.1046 N but I wonder if this is equivalent.
HCl - volume (liters) = 0.01015 L
NaOH - volume (liters) = 0.01 L
2 answers
No, I understand. Thank you very much