They are all an acid/base pair in equilibrium. If they are acidic or basic will depend on how they are prepared.xiao1985 said:
But yes they all will have some effect on the acidity of the solution, but in small amounts it is neglible.
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- Thread starter Budz
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But yes they all will have some effect on the acidity of the solution, but in small amounts it is neglible.
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...
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...
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.
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.
nit
Member
what reaction in particular, xiao?
Budz
Member
Another q's
If 8g of Sulpher dioxide burns completly in air (STP) what volume burnt and what is the mass of water produced?
If 8g of Sulpher dioxide burns completly in air (STP) what volume burnt and what is the mass of water produced?
rama_v
Active Member
S(s) +O2(g) -> SO2(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
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
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Not knowing a whole heap about the prac I guess it works like this;xiao1985 said:
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.