3. 200 ml of 0.2 m hcl is neutralized with 0.1 m
naoh. then during their half neutralization,
what will be the molarity of hci?
a. 0.1 m
b. 0.5 m
c. 0.05 m
d. 1 m

a. 0.1 M

Explanation:

By definition, half neutralization is the point at which half of the acid has been neutralized.

The neutralization reaction that you are studying is the acid-base reaction:

HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)

Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:

Half molarity = M / 2 = 0.2 M / 2 = 0.1 M

So, the answer is the first choice: a. 0.1 M.

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.