A Question A day (1 Viewer)

[helper]

@nightly shadow:
in fact all acid/base indicator are acidic by nature^

They are all an acid/base pair in equilibrium. If they are acidic or basic will depend on how they are prepared.

But yes they all will have some effect on the acidity of the solution, but in small amounts it is neglible.

 

[xiao1985]

10 ml is not quite negligible... =p

this q is inspired by this other day, when i am doing titration... we needed to add 20 ml of starch indicator... i was like "20ml!? that's hell alot"... but since it doesn't change the reaction it doesnt matter as such...

 

[helper]

Agree 10mL isn't normally neglible and you would normally choose the one with the lower level of indicator.

As for the starch indicator. What was the reaction?
As soon as an indicator changes colour, it is an indication it has taken part in a chemical reaction. Depending on the reaction, it may have an effect on the equilibrium or amounts or reactant. The concentration of the indicator is another variable that can effect the results. If its 10mL of a very dilute indicator or 1 mL of concentrated, then the 10mL might be better.

 

[xiao1985]

the reaction involves the oxidation n reduction of I... and starch changes colour when the concetration of I2 (i think) varies...

 

[nit]

what reaction in particular, xiao?

 

[Budz]

Another q's

If 8g of Sulpher dioxide burns completly in air (STP) what volume burnt and what is the mass of water produced?

 

[rama_v]

S_(s) +O_2(g) -> SO_2(g)

8 g of sulphur
Therefore moles of sulphur = 8/32.07 = 0.24945 moles
Moles of sulphur = moles of SO₂ = 0.24945 moles

1 mole of ideal gas occupies volume 24.79 L at STP
Therefore o.24945 mole gas occupies 24.79*0.24945 = 6.184 L

Is that right Im not sure if I interpreted ur question correctly

 

[Jumbo Cactuar]

the reaction involves the oxidation n reduction of I... and starch changes colour when the concetration of I2 (i think) varies...

Not knowing a whole heap about the prac I guess it works like this;

I2 + starch <--> blue complex Kf = low

Solution is blue. Titrate with thiosulfate.
I2 + 2 S2O3 2- --> 2I - + S4O6 2- Kf = high

When near end point think Le Chatelier's principal.
blue complex + 2 S2O3 2- --> 2I- + S4O6 2- + starch Kf = quite high

solution is no longer blue. Hence end point determined. Amount thiosulfate equals half amount I2.