Problems (1 Viewer)

[Aerlinn]

Hm, is this statement, 'a sodium carbonate solution with a concentration of 0.02500M is a primary standard' true/false? I know it's true, but I first had the impression that the primary standard's your pure substance that you dissolve to make the standard solution, not the standard solution itself, or are the two terms interchangeable?

Says somewhere that when you're analysising bleach with a redox titration to find the concentration of the hypochlorite ion, you gotta titrate with sodium thiosulfate, and after the reaction, titrate it with an excess of acidified potassium iodide solution. Curious to know why you'd do two titrations for something like this...

10g of an unknown oil sample is shaken with 100ml of solution containing 15g iodine. The mixture is shaken vigorously and 10.00ml of the unreacted iodine solution is removed and titrated with 0.250M sodium thiosulfate solution. A titure of 21.70ml of sodium thiosulfate is required.
(ans: 8.12g) I did this a couple of times and to my frustration never arrived at the answer >.< (maybe I did something wrong )

In a situation where you get aliquots of household bleach with NaOCl in it, and titrate that with acidified potassium iodide solution. Then you get the free iodine, I2 (a product) and then titrate it with Na2S2O3 using starch indicator solution, which goes blue-black in the presence of iodine... wouldn't the solution turn from blue-black to colourless? The answer says otherwise... says it turns from colourless to blue-black :S

 

[xiao1985]

1) I think it is true... unless they are trying to pull something dodgy like " a primary standard need to be in solid" which I kinda doubt

2) There are lots of ways you can do two titrations in situ. IIRC, the measurement of dissolved O2 is also a 2-3 step titration (Wrinkler method... not the dodgy probe method)..

it can work like this:

A (unknown mole, known volume) + B (known mole, measure the volume) -> C (unknown moles, due to end point non-observable)+ excess B

THEN: either titrate C (unknown moles) + D (known moles)-> E + excess D (end point observabe)
OR excess B + F -> G + excess G (observable end point)

3) The question makes no sense... You don't know what the oil is, so you cannot possibly know what reaction is happening... (ratio, eqn etc)

4) @ end point, I2 (aq) is produced... so the reaction goes something like this: I- (aq) -> I2 (aq) and it's the I2 (aq) which makes the solution turn blue/black, not I-

 

[Aerlinn]

Ah, ok...

3)Ok, this is the background to the question, the stuff that goes before it (didnt realise it was needed, sowwi)
Unsaturated oils contain one or more double cabon to carbon bonds. Iodine reacts readily with these double bonds:
Chemists measure the degree of unsaturation by determining the iodine number of the oil. The iodine number is the no. grams of iodine that reacts with 100g of oil. The oil is put in a stoppered flask, a known excess of iodine is added and the flask shaken to allow the iodine to reaact. the reamining iodine is determined by titration with sodium thiosulfate solution.

4) And this is the whole question for this one.
Liquid household bleach contains sodium hypcholorite (NaOCl) as its active ingredient. The manufacturer maintains quality control by performing volumetric analysis on random samples. In the analytical laboratory a 25.00ml sample was diluted to 250.0ml with distilled water. 20.00ml aliquots were pipetted into conical flasks and excess acidified potassium iodide solution was added to each flask to cause the reaction:
ClO-(aq) + 2I-(aq) + 2H+(aq)----> Cl-(aq) + I2(aq) + H20(l)
The free iodine was then titrated with 0.1000M sodium thiosulfate (na2S2O3) using starch indicator solution. This indicator givesa blue-black colour in the presence of iodine
The titration reaction is r epresented by:
2S2O3(2-) (aq) + I2(aq) ---> S4O6(2-) +2I-(aq)
Dunno why I typed it out, but if they added the starch in the the second reaction, and its blue black for I2, wouldnt it go blue-black ---> clear?!
Dun understand

 

[xiao1985]

thiol sulphate react with excess I2 in 1:1 fashion:

S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
(http://en.wikipedia.org/wiki/Thiosulphate)

0.25 x 0.0217 = 0.005425 moles of thiosulphate reacted = 0.005425 moles I2 left
= 0.005425 x (126.9 x 2) = 1.376865 g of I2 left = 15-1.376865 = 13.62 g reacted = 13.62/(126.9x2) = 0.0536 moles of I2 reacted...

hmm... ok... so what's the question asking really? i figured out as much as i could, but i still don't understand the question...

q4: oh well... perhaps it was I- which gives the purple color... sorry for the misinformation...
(http://en.wikipedia.org/wiki/Starch ; note detect iodide ion... not iodine)

 

[Aerlinn]

For that first one, *wacks self over head* You can tell I wasn't thinking properly can't you? I was stupid enough to leave out the actual question, that's why (I give you permission to wack me ^^)

Anyway, the WHOLE question this time.
Unsaturated oils contain one or more double cabon to carbon bonds. Iodine reacts readily with these double bonds:
Chemists measure the degree of unsaturation by determining the iodine number of the oil. The iodine number is the no. grams of iodine that reacts with 100g of oil. The oil is put in a stoppered flask, a known excess of iodine is added and the flask shaken to allow the iodine to reaact. the reamining iodine is determined by titration with sodium thiosulfate solution.

10g of an unknown oil sample is shaken with 100ml of solution containing 15g iodine. The mixture is shaken vigorously and 10.00ml of the unreacted iodine solution is removed and titrated with 0.250M sodium thiosulfate solution. A titure of 21.70ml of sodium thiosulfate is required.

First up: Overall equation for the reaction between iodine and sodium thiosulfate, given that the thiosulfate ion (s2O3(2-)) is oxidised to S4O6(2-)) and I2 is reduced to I-
So: I2(aq) + 2S2O3(2-)(aq) ---> 2I-(aq) + S4O6(2-)(aq)
Then: calculate the no. moles of iodine the the 10 ml aliquot, which is 2.71*10^-3 mol.
THEN Calculate the number of grams of iodine that reacted with the 10g of oil. (this is the actual question)
(ans: 8.12g)
~~~
Ah, iodine ions. That does make more sense now

 

[xiao1985]

yup it does... and there is an error in my previous working out...

you mean iodide ions? or iodine molecules?

eh? how come i got 0.005425 moles of Iodine ? hmm... i think i know what's wrong... you are asked about iodine (I2), not iodide(I-), and since thiosulfate and I2 react in 1:1 ratio, the number of moles of I2 present is 0.25 * 0.0217 = 0.005425 moles.

there is 0.005425 moles of I2 left in 10mL of the aquilot... which means, in the total 100mL, there are 0.05425 moles of I2 left = 13.77g of I2 unreacted...

reacted = 15-13.77 = 1.23 g?????

errr ^^''