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. 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

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

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#2

Okay so you’ve worked out how many moles of iodine were produced.

Using the first equation what can this also tell you

Using the first equation what can this also tell you

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(Original post by

Okay so you’ve worked out how many moles of iodine were produced.

Using the first equation what can this also tell you

**GabiAbi84**)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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#4

(Original post by

that 1.025x10-3 moles of ClO- ions reacted since it's a 1:1 ratio?

**CJ_bangtan**)that 1.025x10-3 moles of ClO- ions reacted since it's a 1:1 ratio?

So now you know number of moles in 25cm3

What is your number of moles for 250cm3?

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(Original post by

Good,

So now you know number of moles in 25cm3

What is your number of moles for 250cm3?

**GabiAbi84**)Good,

So now you know number of moles in 25cm3

What is your number of moles for 250cm3?

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#6

(Original post by

would it be 4.1x10-4 as you multiply by 10

**CJ_bangtan**)would it be 4.1x10-4 as you multiply by 10

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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(Original post by

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)

**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)

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#8

(Original post by

ahhh i see so it would be 0.01025 moles

**CJ_bangtan**)ahhh i see so it would be 0.01025 moles

Now do you know what to do from there?

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#10

(Original post by

do you now need to find the number of moles in 10cm3 and use that value to find the concentration by doing moles/volume?

**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?

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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(Original post by

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...

**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...

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#12

(Original post by

4.1x10-4 moles?

**CJ_bangtan**)4.1x10-4 moles?

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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(Original post by

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.

**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.

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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#14

(Original post by

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

**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

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(Original post by

Yes, perfect!

**GabiAbi84**)Yes, perfect!

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#16

(Original post by

so it would be 0.01025/10 to find the concentration

**CJ_bangtan**)so it would be 0.01025/10 to find the concentration

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(Original post by

Remember that volume needs to be in dm3 not cm3

**GabiAbi84**)Remember that volume needs to be in dm3 not cm3

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#18

(Original post by

. 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

**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

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#19

(Original post by

. 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

**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

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#20

(Original post by

Where did you find this questions? Can you please dm me the original question paper?

**1025200275**)Where did you find this questions? Can you please dm me the original question paper?

The question is all there...

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