# back titration

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#1
. Sodium hypochlorite NaClO is the active ingredient in bleach. Cocentration of the hypochlorite ion determined in two ways
Stage 1 acidified iodide solution added to a solution of bleach and iodine formed
ClO- + 2I- + 2H+ = I2 + Cl- + H2O
Stage 2 iodine formed titrated with sodium thiosulfate solution
2S2O3 + I2 = 2I- + S4O6 2-
10cm3 of a household bleach diluted to 250cm3 in standard flask
25cm3 of this solution added to excess scidified potassium iodide solution
Solution the titrated against 0.10mol l-1 sodium thiosulfate
The volume of the thiosulfate required to reach end point was 20.5cm3
A) Calculate moles of iodine reacted in the titration
B) Calculate the concentration in mol l-1 of the ClO- in original household bleach
for part A I got 1.025x10 -3 moles
for part B I got 1.64x10-3 mol l-1
Could anyone tell me if this is the right answer because I can't find the answers to this question
Last edited by CJ_bangtan; 3 weeks ago
0
3 weeks ago
#2
Okay so you’ve worked out how many moles of iodine were produced.
Using the first equation what can this also tell you
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#3
(Original post by GabiAbi84)
Okay so you’ve worked out how many moles of iodine were produced.
Using the first equation what can this also tell you
that 1.025x10-3 moles of ClO- ions reacted since it's a 1:1 ratio?
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3 weeks ago
#4
(Original post by CJ_bangtan)
that 1.025x10-3 moles of ClO- ions reacted since it's a 1:1 ratio?
Good,
So now you know number of moles in 25cm3
What is your number of moles for 250cm3?
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#5
(Original post by GabiAbi84)
Good,
So now you know number of moles in 25cm3
What is your number of moles for 250cm3?
would it be 4.1x10-4 as you multiply by 10
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3 weeks ago
#6
(Original post by CJ_bangtan)
would it be 4.1x10-4 as you multiply by 10
Your number of moles of ClO- in 25cm3 is 1.025x10-3
So yes you multiply that by 10 to get number of moles in 250.

(But it isn’t 4.1x10-4)
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#7
(Original post by GabiAbi84)
Your number of moles of ClO- in 25cm3 is 1.025x10-3
So yes you multiply that by 10 to get number of moles in 250.

(But it isn’t 4.1x10-4)
ahhh i see so it would be 0.01025 moles
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3 weeks ago
#8
(Original post by CJ_bangtan)
ahhh i see so it would be 0.01025 moles
Yes.
Now do you know what to do from there?
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#9
(Original post by GabiAbi84)
Yes.
Now do you know what to do from there?
do you now need to find the number of moles in 10cm3 and use that value to find the concentration by doing moles/volume?
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3 weeks ago
#10
(Original post by CJ_bangtan)
do you now need to find the number of moles in 10cm3 and use that value to find the concentration by doing moles/volume?
Yes.

But think carefully about the number of moles in that 10cm3 - remember it’s just been diluted so the number of moles will be...
Last edited by GabiAbi84; 3 weeks ago
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#11
(Original post by GabiAbi84)
Yes.

But think carefully about the number of moles in that 10cm3 - remember it’s just been diluted so the number of moles will be...
4.1x10-4 moles?
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3 weeks ago
#12
(Original post by CJ_bangtan)
4.1x10-4 moles?
Not sure where you’re getting that from.

When you dilute a solution you just add water. So the number of moles doesn’t change, just the volume and concentration does.
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#13
(Original post by GabiAbi84)
Not sure where you’re getting that from.

When you dilute a solution you just add water. So the number of moles doesn’t change, just the volume and concentration does.
so it would just be 0.01025
The reason I got 4.1x10-4 was because I thought that since 250cm3 contains 0.01025moles then 10cm3 would contain 4.1x10-4 moles
so the number of moles doesn't change because since your adding water and diluting the solution your decreasing the concentration and increasing the volume and the moles of a substance is spread out over a larger volume therefore doesn't change
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3 weeks ago
#14
(Original post by CJ_bangtan)
so it would just be 0.01025
The reason I got 4.1x10-4 was because I thought that since 250cm3 contains 0.01025moles then 10cm3 would contain 4.1x10-4 moles
so the number of moles doesn't change because since your adding water and diluting the solution your decreasing the concentration and increasing the volume and the moles of a substance is spread out over a larger volume therefore doesn't change
Yes, perfect!
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#15
(Original post by GabiAbi84)
Yes, perfect!
so it would be 0.01025/10 to find the concentration
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3 weeks ago
#16
(Original post by CJ_bangtan)
so it would be 0.01025/10 to find the concentration
Remember that volume needs to be in dm3 not cm3
1
#17
(Original post by GabiAbi84)
Remember that volume needs to be in dm3 not cm3
ok I think I finally understand how to do this question now thank you so much 0
5 days ago
#18
(Original post by CJ_bangtan)
. Sodium hypochlorite NaClO is the active ingredient in bleach. Cocentration of the hypochlorite ion determined in two ways
Stage 1 acidified iodide solution added to a solution of bleach and iodine formed
ClO- + 2I- + 2H+ = I2 + Cl- + H2O
Stage 2 iodine formed titrated with sodium thiosulfate solution
2S2O3 + I2 = 2I- + S4O6 2-
10cm3 of a household bleach diluted to 250cm3 in standard flask
25cm3 of this solution added to excess scidified potassium iodide solution
Solution the titrated against 0.10mol l-1 sodium thiosulfate
The volume of the thiosulfate required to reach end point was 20.5cm3
A) Calculate moles of iodine reacted in the titration
B) Calculate the concentration in mol l-1 of the ClO- in original household bleach
for part A I got 1.025x10 -3 moles
for part B I got 1.64x10-3 mol l-1
Could anyone tell me if this is the right answer because I can't find the answers to this questionF
From which past paper is this question?
0
5 days ago
#19
(Original post by CJ_bangtan)
. Sodium hypochlorite NaClO is the active ingredient in bleach. Cocentration of the hypochlorite ion determined in two ways
Stage 1 acidified iodide solution added to a solution of bleach and iodine formed
ClO- + 2I- + 2H+ = I2 + Cl- + H2O
Stage 2 iodine formed titrated with sodium thiosulfate solution
2S2O3 + I2 = 2I- + S4O6 2-
10cm3 of a household bleach diluted to 250cm3 in standard flask
25cm3 of this solution added to excess scidified potassium iodide solution
Solution the titrated against 0.10mol l-1 sodium thiosulfate
The volume of the thiosulfate required to reach end point was 20.5cm3
A) Calculate moles of iodine reacted in the titration
B) Calculate the concentration in mol l-1 of the ClO- in original household bleach
for part A I got 1.025x10 -3 moles
for part B I got 1.64x10-3 mol l-1
Could anyone tell me if this is the right answer because I can't find the answers to this question
Where did you find this questions? Can you please dm me the original question paper?
0
5 days ago
#20
(Original post by 1025200275)
Where did you find this questions? Can you please dm me the original question paper?
Why do you need to know?
The question is all there...
0
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