Question

II Titrating the Diluted HCI Solution normality of NaOH solution 0.1023 () = 0.1023 volume of diluted HCl solution titrated, mL N- 25.00 molarity HCI: O.IM molarity NaOH-0.1023 m determination final buret reading, mL initial buret reading, mL 24.4 38.8 244 26,17 35,2 0.2 144 o 2,6 1668 24.2 24.4 24.4 24 244 UUUUU volume of titrant, ml 99999SSSSSSSSSS normality of diluted HCI solution, equiv L- mean normality of diluted HCl solution, equiv L-1 UUUUUUUU normality of concentrated HCl solution, equiv L-

I need to know how to calculate the normality of these. help is greatly appreciated!!!

Answer 1

The equivalents of the NaOH and HCl are equal at the end point, since we re titrating the diluted HCl and the NaOH, thus,

NV (diluted HCl) = NV (NaOH)

Using this equation, we can calculate the Normality of diluted HCl for the different trials.

For Trail 1, N (NaOH) = 0.1023

V (NaOH) = 24.2ml (burette rading)

V (diluted HCl) = 25ml

Thus N (diluted HCl) = 0.1023*24.2/25 = 0.0990N

Similarly for trail 2,

N (diluted HCl) = 0.1023*24.4/25 = 0.0998N

Similarly for trial 3,

N (diluted HCl) = 0.1023* 24.4/25 = 0.0998N

Similarly for trial 4,

N (diluted HCl) = 0.1023*24.1/25 = 0.0986N

Similarly for trial 5,

N (diluted HCl) = 0.1023 * 23.4/25 = 0.0957N

Therefore, mean Molarity of Diluted HCl solution

=> (0.0990+ 0.0998 + 0.0998 + 0.0986 + 0.0957)/5

=> 0.0986N

Again, applying the equation of equal equivalents for diluted and concentrated HCl,

NV (Diluted HCl) = NV (Concentrated HCl)

We have, N (diluted HCl) = 0.0986 N

V (diluted HCl) = 25ml

So if know the V of concentrated HCl, we can calculate the N of Concentrated HCl.