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October 21, 2025, 10:34:03 pm

Author Topic: Titration practical exam!  (Read 4727 times)  Share 

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ArsenalFanTv

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Titration practical exam!
« on: May 14, 2017, 09:42:33 pm »
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Hey Guys,
I'm not so sure where to start studying for my titration prac exam. I was wondering if anyone has any idea of what I would expect and what I require to know.
Thanks!

jakesilove

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Re: Titration practical exam!
« Reply #1 on: May 14, 2017, 09:45:14 pm »
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Hey Guys,
I'm not so sure where to start studying for my titration prac exam. I was wondering if anyone has any idea of what I would expect and what I require to know.
Thanks!


Hey! Check out the below tips from an old user of ours :)

Hello Neeshy:

Yes I can definitely understand how those titration prac exams can be extremely concerning, but dont worry, we are all here to help you! I will provide you with some specific tips below and hope they help.

Preparation of a Primary Standard Solution:
- A standard solution is a solution of known composition and concentration used for quantitative analytical procedures (in this case, titration)
- Standard solution made by dissolving a primary standard volume of water
- Solids substances are desirable for preparing standard solution because they are easier to weigh
- It is also desirable for the substances to be easily obtained in its pure form, which reduces the cost and makes it easier to perform calculations with minimal errors involving the existence of impurities
- Substances with large molar mass is desirable to allow for reduced weighing errors
- And finally, the substance must also have a high water solubility - obviously, because the substance is required to be dissolved in water for standard solution preparation

*Procedure for preparing primary standard solutions may be obtained from your textbook

Dilution Calculations

- Since we are adding only solvent, moles of solute remains unchanged
- Before we dilute (solution 1, concentrated) the equation that applies is n1 = C1V1
- After we dilute, V and C have both changed, hence the equation that applies is n2=C2V2
- But since the moles of solute is the same before and after, hence the equation that applies is C1V1 = C2V2

General Overview of Neutralisation Reaction

- HEAT is released in neutralisation reactions between strong acids and strong bases (e.g. NaOH + HCl, the most common example)
- ACID + BASE ----> SALT + WATER
- Neutralisation reaction between strong acid and strong base is essentially the reaction of a proton and OH ion to form water

Titration

- An analytical process which involves finding the volume of a solution of accurately known conc. that is required to react completely with a known volume of a solution of unknown concentration. Hence it is a volumetric analysis.
- Summary of Procedure:
       1. A known volume of a solution of unknown conc. placed in a conical flask
       2. A solution of known conc. added until all of the analyte has reacted, volume is recorded
       3. Using the mole ratio of the reaction the unknown conc. is calculated

Calculation of Concentration of an Unknown Solution:

1. Write a balanced equation
2. Calculate number of moles of reactant A (n=CV)
3. Using the moles of reactant A and the stoichiometric molar ratio in the equation, calculate the number of moles of Reactant B used
4. Use C=n/V to calculate the conc. of Reactant B

These are basically all I have to say, a bit content heavy, if Jake happens to come around he can give you more advices. Anyways good luck my friend, if you have any questions, dont hesitate to ask!

Best Regards
Happy Physics Land

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