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Topic: Volumetric analysis  (Read 14972 times)

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chemistrie

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Volumetric analysis
« on: December 20, 2004, 10:15:35 PM »
I have a few questions about this topic:

1. Why does the molarity of a solution is defined as "number of moles of solutes in a certain volume of solution" rather than volume of water added?

2. When preparing 1L standard solution, the solute is first dissolved in water and then made up in the volumetric flask. Why is the solute not simply dissolved in 1L of water?

3. Can I directly dissolve the solid directly inside the volumetric flask rather than adding certain amount of water to a beaker and dissolve the solid first before pouring all the washings into the flask?

4. During titrations, we are advised to wash the pipette and burette with solutions they are going to contain instead with only water, why? And we are told it's not necessary to dry the conical flask after rinsing with water, but won't it affect the molarity of the solution it's going to contain?

Offline AWK

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Re:Volumetric analysis
« Reply #1 on: December 21, 2004, 01:59:52 AM »
1. This definition comes from experience.
1,2 - read somethin on "contraction of volume " in Physical Chemistry textbooks
3. It is better to follow procedure
4. During titration you compare number of moles of two reagents, so you should have exact concentration in burette (moles=concntration times volume in liters), and exact moles of  titrated reagent (not necessrily its concentration).
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Offline kevins

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Re:Volumetric analysis
« Reply #2 on: December 21, 2004, 03:22:17 AM »
chemistrie,

For Q3.
Basically it is work, but one thing should bear in mind that no volume change( contraction and expansion when solute dissolve) and no heat given out ( the glass will expand and the exact volume change, so a temperature 20C is mark on the glass which means the volume of the flask is calibrated with this temperature of water) during the top up of the volumetric flask. Therefore, on the safe side, the preparation procedure for titration would state that "to dissolve the solute in a beaker with little amount of solvent".Right?

For Q4.
If little amount of water left in the burette or in the pipette, the concentration of the solution will change. Right? Then the error come out.
The water in the conical flask does not affect the result. Do you know why?

chemistrie

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Re:Volumetric analysis
« Reply #3 on: December 21, 2004, 03:41:00 AM »
But if water is present inside the conical flask, isn't the volume of the solution will be increased, hence causing the molarity of the solution becomes lower?

Offline kevins

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Re:Volumetric analysis
« Reply #4 on: December 21, 2004, 09:42:03 AM »
Yes, the volume of solution is change. How about the amount (not the concentration) of the reactant? any change?

chemistrie

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Re:Volumetric analysis
« Reply #5 on: December 21, 2004, 10:15:24 PM »
But if we view that water inside the conical flask do nothing to change the concentration of the solution , then why we can ignore the effects of water inside the flask but not the solution inside the burette? If the number of moles of solute inside the conical flask remains unchanged even though water is being added, then how come the concentration of solution inside the burette will change? Isn't the number of moles of solutes inside the burette also remains unchanged?

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Re:Volumetric analysis
« Reply #6 on: December 22, 2004, 01:39:34 AM »
compare these 2 examples:

1L of Liquid A combines with 1L of liquid B to produce 1.7L Mixture

Add 1L of Liquid A inside a standard flask. Top up the flask with Liquid B until the 2L mark.
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chemistrie

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Re:Volumetric analysis
« Reply #7 on: December 22, 2004, 04:32:52 AM »
We should have exact concentration in burette and exact moles of titrated reagent , whose concentration is not known, but not necessrily its concentration, why?

And Chemist kevins stated that the amount of water inside the conical flask does not affect concentration of the solution, but water inside the burette or pipette will, how come?

Offline Mitch

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Re:Volumetric analysis
« Reply #8 on: December 22, 2004, 04:40:10 AM »
4.) If you've done the experiment, you will know the volume difference on the burette is what matters not the "conical" flask..
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chemistrie

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Re:Volumetric analysis
« Reply #9 on: December 22, 2004, 05:52:43 AM »
Then we should not wash the conical flask with the solution it is going to contain. Why?

Can anyone restate why water inside the burette affects the molarity of solution but water inside the conical flask doesn't have this problem? I still don't quite understand and get the concept.

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Re:Volumetric analysis
« Reply #10 on: December 22, 2004, 01:27:06 PM »
Have you done this experiment? It really isn't that complicated if you have done it.
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Demotivator

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Re:Volumetric analysis
« Reply #11 on: December 22, 2004, 03:26:14 PM »
Let me put it this way.
In the burette, you are measuring volume delivered. If the burette is not dry, the extra volume will reduce the concentration as you have noted, and therefore affect the totat volume required for the end point.

But in the conical flask, you know the volume you introduced into the flask right? For, eg. if I know I put in 50 ml of solution into a wet conical flask, no matter how much water I add to it, I still know the original volume was 50 ml. Any extra water will be excluded from the calculation. The titration will tell me total moles. The conc in flask is then tot moles/50 ml.

chemistrie

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Re:Volumetric analysis
« Reply #12 on: December 23, 2004, 01:02:55 AM »
That means the volume of the solution inside the burette is not known?

Offline Mitch

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Re:Volumetric analysis
« Reply #13 on: December 23, 2004, 02:44:29 AM »
The point of the experiment is to calculate the volume as accurately as possible.
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chemistrie

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Re:Volumetric analysis
« Reply #14 on: December 23, 2004, 03:40:48 AM »
Suppose there's 1M NaOH solution and unknown concentration of HCl whose volume is 30mL. If water is present and the volume of HCl increased to 40mL. If no water is present inside the flask, suppose 25mL NaOH completely reacts with the acid. But if water is present, would it need extra volume of NaOH to reach the end point? say, 30mL? Then the experimental result would then be affected?

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