Question

Suppose a student diluted and titrated a bleach unknown. A single titration was performed which required 14.42 mL of 0.100 M Na2S2O3.
The density of the original, undiluted bleach unknown was 1.042 g/mL.

calculate the number of moles of ClO- in the sample titrated.

Answer 1

You're post is too sketchy. For example, did you add KI to the bleach to free up I2 and the I2 was titrated with thiosulfate? Of did you use some other procedure.

Answer 2

sorry this is for a prelab. But yes.

Answer 3

ClO^- + 2I^- ==> I2 + Cl^- then the titration step is
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-

1 mol ClO^- = 1 mol I2
1 mol I2 = 2 mol S2O3^2-
therefore, 1 mol ClO^- = 2 mol S2O3^2- or
1/2 mol ClO^- = 1 mol S2O3^2-

mols S2O3 = M x L
mols ClO^- = 1/2 that.

Answer 4

thank you

Answer 5

one last question.
From same set up.

Assuming all of the hypochlorite ion comes from sodium hypochlorite, calculate the grams of NaClO in the titrated bleach sample.

Answer 6

mols ClO^- x molar mass NaClO = ?

Answer 7

5 mL of undiluted bleach was used. Calculate the mass in grams of undiluted bleach that was in the sample of bleach titrated.

Answer 8

The same answer as previous.
mols ClO^- x molar mass NaClO = grams NaClO.

If you want percent then
(g bleach/g sample)*100 = ?
g bleach from above.
g sample = 5 mL x density g/mL = grams sample

Answer 9

use this to get the mass of the bleach? It is telling me my answer is incorrect.

Answer 10

If you care to post your work I will be glad t check it. Don't take shortcuts in doing it. When checking problems against an on-line database the problem (and it happens quite often) is you key in too many (or not enough) significant figures. Check that first before typing in your work.

Answer 11

No where do I see anything in your post that relates to the 5 mL sample titrated. Re-read the problem and see what the original sample was then apply the 5 mL aliquot to it. I suspect that's the error.

Answer 12

to get the grams of NaClO in the titrated bleach sample I calculated .000721 x 74.44217 = .0536 which was correct.

For the previous question I asked I put in the same answer as you said and it is incorrect.

Answer 13

For 0.01442 x 0.100 x 1/2 x 74.442 = 0.0537 g NaClO. From your description I can't tell the difference between this one and the one with 5 mL in it. The only difference in the two questions is the 5 mL and you have no information in any of the problem that relates to the 5 mL. So I've not used that. This problem often happens when ALL of the problem is not typed.