April 21, 2019, 04:27:41 PM
Forum Rules: Read This Before Posting

### Topic: REDOX and Iodometry *delete me*  (Read 2164 times)

0 Members and 1 Guest are viewing this topic.

#### twomblyhero

• New Member
• Posts: 7
• Mole Snacks: +0/-0
##### REDOX and Iodometry *delete me*
« on: June 18, 2014, 05:45:03 AM »
We completed a redox titration of Iodine (25cm3)
(formed from 1M Potassium Iodide (KI), 0.1M Potassium Iodate (KIO3) and excess Hydrochloric acid (KCl)

against Sodium thiosufate (Nas2O3)

It took 25.8cm3 of 0.1 M Nas2O3 to get rid of the formed Iodine.

Use the equation below to calculate the amount of iodide ions that reacts with each mole of iodate ions in aqueous solution

2S2O32- (aq)+ I2 (aq)     2I- (aq) + S4O62-(aq)

and determine the concentration of iodide ions in moldm-3

I am not sure what this is asking me! Do I need to calculate how the Iodine is originally formed?

« Last Edit: June 18, 2014, 08:07:23 AM by twomblyhero »

#### Hunter2

• Sr. Member
• Posts: 1762
• Mole Snacks: +130/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: REDOX and Iodometry *delete me*
« Reply #1 on: June 18, 2014, 06:39:41 AM »
They want to know how much iodide was in the solution.

You have a sample of Iodide it reacts with iodate to iodine. Develop here the equation as well. The iodine is titrated. So you can calculate back.

#### twomblyhero

• New Member
• Posts: 7
• Mole Snacks: +0/-0
##### Re: REDOX and Iodometry *delete me*
« Reply #2 on: June 18, 2014, 07:32:24 AM »

I know
5KI + KIO3 + 6HCL  3H2O + 6KCl + 3I2

and the ratio of 2S2O32- (aq)+ I2 (aq) is 2:1 I can work out the concentration of I2 but then how do I work the rest out?

Any ideas folks?

#### Hunter2

• Sr. Member
• Posts: 1762
• Mole Snacks: +130/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: REDOX and Iodometry *delete me*
« Reply #3 on: June 18, 2014, 07:41:36 AM »
Simple mathematics. You see how much iodide creates iodine.  And youn see how much thiosulfate is used for Iodine. You have to calculate the relation Iodide Thiosulfate.

#### twomblyhero

• New Member
• Posts: 7
• Mole Snacks: +0/-0
##### Re: REDOX and Iodometry *delete me*
« Reply #4 on: June 18, 2014, 07:51:58 AM »
So 5 moles of Iodide ions (from KI) will react with 1 mole of Iodate ions (from KIO3)?

concentration= moles         =  5   x 25/1000 = 0.0002mols dm-3
_____
volume

....still stuck!

#### Hunter2

• Sr. Member
• Posts: 1762
• Mole Snacks: +130/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: REDOX and Iodometry *delete me*
« Reply #5 on: June 18, 2014, 09:09:12 AM »
5 mol iodide create 3 mole iodine.

2 mol  thiosulfate correspond to 1 mol iodine

how many mole iodide correspond to thiosulfate?

#### twomblyhero

• New Member
• Posts: 7
• Mole Snacks: +0/-0
##### Re: REDOX and Iodometry *delete me*
« Reply #6 on: June 18, 2014, 09:22:05 AM »
1?

#### Hunter2

• Sr. Member
• Posts: 1762
• Mole Snacks: +130/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: REDOX and Iodometry *delete me*
« Reply #7 on: June 18, 2014, 09:24:43 AM »
No it is 6 mol Thiosulfate correpond to 5 mol Iodide.

With his knowledge calculate how many mole 25.8 ml of 0,1 M thiosulfate is.  If you have this then you can calculate the content of iodide. This you have to get in relation to your 25 ml sample.

#### twomblyhero

• New Member
• Posts: 7
• Mole Snacks: +0/-0
##### Re: REDOX and Iodometry *delete me*
« Reply #8 on: June 18, 2014, 09:36:11 AM »
Thanks for you help, I understand the ratios now. Just got to do the calculations now.