blackmamba123
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Dilution problem
Hey guys I have a question:
How would I find the amount of diluted IO3- given the following information:
-10 ml of IO3- is added to 10ml of bisuplhate ions
- Assuming that the IO3- concentration is 0.001 M before any mixing occurs.
I have tries using the formula C1V1=C2V2, but i just don't understand the concept.
Any help is appreciated.
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aga
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Surely the bisulphate ions would need some solvent to be ions in ?
i.e. the Ions would be a certain [M] of ions in, say, 10ml of water, not just 10ml of ions.
[Edited on 18-5-2015 by aga]
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blogfast25
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Quote: Originally posted by blackmamba123  | Hey guys I have a question:
How would I find the amount of diluted IO3- given the following information:
-10 ml of IO3- is added to 10ml of bisuplhate ions
- Assuming that the IO3- concentration is 0.001 M before any mixing occurs.
I have tries using the formula C1V1=C2V2, but i just don't understand the concept.
Any help is appreciated. |
You need to specify which reaction you are trying to achieve, w/o this information the question is literally impossible to answer. Also, If you're not
to achieve a reaction (also possible) then what is your objective?
Then I'll show you how to use the C<sub>1</sub>V<sub>1</sub>=C<sub>2</sub>V<sub>2</sub> rule.
Aga, he's (not too) clearly referring to a solution of bisulphate, presumably in water.
[Edited on 18-5-2015 by blogfast25]
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blackmamba123
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Sorry about not being specific.
The question is asking for the amount of diluted IO3- when the two substances are mixed. ( The solution of IO3- and the solution of bisulphate ions,
acid and starch.)
I also know that the total volume when the two substances are mixed together is ml, since both had an initial volume of ml before being mixed.
Hope I'm being specific enough
[Edited on 18-5-2015 by blackmamba123]
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aga
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Yup.
You can't have a jar of ions or cations.
It will be a Molar concentration of ions in a Solution, i.e. 10ml of water with a [M] 0.001 of IO3- in that solution
That would be your C1 (0.001) and V1 (10ml).
If that's added to 10ml of another solution then that's your V2 (10ml + 10ml).
You only really get ions in a solution, e.g. Table Salt is NaCl, and dissolves in water (the solvent) to make the ions Na+ and Cl-.
The concentration (C1) is how many ions, and the volume is how many mls of water (V1).
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blackmamba123
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from that how would i figure out the amount of diluted IO3- once the two substances are mixed?
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aga
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C1V1 = C2V2
You already got:-
C1 = 0.001 (molar concentration of IO3- ions)
V2 = 10ml (of the solvent those ions were dissolved in)
They got added to 10ml of another solvent, containing bismuth ions.
How much Volume is there now ?
10ml + 10ml = 20 ml, and this is your V2.
Re-arrange the formula to get your C2 on it's own :
C2 = (C1V1) / V2
So C2 = (0.001 x 10 ) / 20 = ......
Edit:
The concept is this:-
If you have X molecules of something in Y volume of a solvent (say Water) then if you Reduce the amount of water, the Concentration of the stuff must
Increase.
If you Increase the amount of water, the Concentration must Decrease.
So C1V1 = C2V2 where C1V1 is the concentration of molecules in the starting volume of solvent multiplied by the original volume of Solvent, and C2 is
the concentration of those molecules if you change the volume (i.e. add more water) to change V1 into V2.
Basically, add water to your soda and there is a less concentrated flavour.
[Edited on 18-5-2015 by aga]
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blogfast25
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| Quote: Originally posted by aga | C1V1 = C2V2
You already got:-
C1 = 0.001 (molar concentration of IO3- ions)
V2 = 10ml (of the solvent those ions were dissolved in)
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You're making this unnecessarily complicated. The indices always refer to 'before' (State 1) and 'after' (State 2). Don't mix it up, it will confuse
your students  .
This is a simple case of dilution. 10 ml of a solution is diluted by adding 10 ml of another solution. No reaction takes place.
Before dilution (State 1): c1 = 0.001 M, volume is V1 = 10 ml.
After dilution (State 2): c2 (unknown), volume is V2 = V1 + 10 ml = 20 ml.
c1 x V1 = c2 x V2
c2 = c1 x (V1/V2) = 0.001 M x 10/20 = 0.0005 M
Note that the actual AMOUNT of IO<sub>3</sub><sup>-</sup> in moles hasn't actually changed, it's only contained in twice its
volume after the dilution.
BISULPHATE is not BISMUTH, BTW aga. 
[Edited on 18-5-2015 by blogfast25]
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blackmamba123
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Thank you so much! I understand the concept now.. So all I have to do is solve for the C2, correct?
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aga
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Yes.
The whole C1V1 = C2V2 thing is to work out the Concentration of your stuff, Before and After you dump it in More water (or another solvent).
e.g.
You get 58.44g of pure table salt (look up the molecular mass of NaCl).
You throw it into a bucket with 1 litre of water (V1 = 1litre)
Your concentration is 1 mole of NaCl in 1 Litre of water = 1 [M] (C1 = 1)
Add another Liter of water. Total Volume is now 2 Litres (V2 = 2 Litres).
C1V1 = C2V2
So the concentration of salt is obviously Half what it was to start with.
Check for yourself.
C1V1 = C2V2
C2 = ( 1 x 1 ) /2 = ...
[Edited on 18-5-2015 by aga]
[Edited on 18-5-2015 by aga]
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blogfast25
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Yes but always attributing the correct indices of c and V to 'before' and 'after'.
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aga
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I was obviously not Concentrating.
My right C2 just got smaller !
Such is the power of a transitional Jedi/Sith master when castigating careless students.
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blogfast25
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| Quote: Originally posted by aga |
I was obviously not Concentrating.
My right C2 just got smaller !
Such is the power of a transitional Jedi/Sith master when castigating careless students. |
An investigation of your teacher has now been initiated. 'Heads must surely roll shortly?' one observer was heard asking. 'His flawless CV won't save
him now!' observed another.
[Edited on 18-5-2015 by blogfast25]
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aga
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Head / Chopper.
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blogfast25
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I wouldn't worry your pretty head over it, Marie Antoinette!
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blackmamba123
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From dilution problems to Marie Antoinette, how did this unravel?
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aga
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Welcome to ScienceMadness !
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