How to prepare 0.1 M of NaOH for Titration lab

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How to prepare 0.1 M of NaOH for Titration lab

Postby Traumatizes » Mon Jan 14, 2013 4:30 pm

I have to prepare 500 mL of 0.1 M NaOH. Please can someone tell me, step by step how to this this. It is for a lab and very important.
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Project Question: For a chemistry project, I have chosen to measure the Amount of Acid in Vinegar by Titration with an Indicator Solution. I have chosen 3 types of vinegar for this project; White, Apple Cider, and Malt. To titrate, I need 0.1 M of NaOH and I need of enough for 3 trials per vinegar (So 9 trials altogether) My question is, how do you get 0.1 M NaOH? I need around 500 mLs (450 to be exact but 500 would be ideal)
Project Due Date: We only get 2 more days of lab time so all experiments are to be done by Wednesday, January 16, 2013. Then we have until Monday, January 21, 2013 to have our lab report done.
Project Status: I am conducting my experiment

Re: How to prepare 0.1 M of NaOH for Titration lab

Postby Ray Trent » Mon Jan 14, 2013 10:07 pm

Here is a web site showing how to prepare various solutions for chemistry experiments. You'll need to find some NaOH in solid form and a chemical scale. I would imagine both would be in your lab, as they are common items.

http://abacus.bates.edu/~ganderso/biolo ... tions.html
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Re: How to prepare 0.1 M of NaOH for Titration lab

Postby billeykamp » Thu Jan 17, 2013 11:02 am

I just answered a similar question for someone else. Perhaps it will be helpful to you. Here goes. Good luck.

Well, I did this often 50 years ago. Let's hope someone more up-to-date will weigh in. If not, here goes.

You say 0.1M NaOH. I will speak in equivalents, not molarity. Don't despair, for NaOH it's the same. For other things it's not, and equivalents is far more general, and is usually used. Acetic acid also has equivalent weight and molar mass the same. "N" stands for normal.

Potassium hydrogen phthalate is considered a stable, reliable chemical for calibrating other acids and bases. It has a molar mass of 204.2 (check my result yourself, please) and has an equivalent weight equal to its molar mass. That means there will be one reactable hydrogen ion in every molecule. Sodium hydroxide, whose molar mass is about 40, also has an equivalent weight equal to its molar mass. It has one reactable hydroxyl ion per molecule.

The equipment I'd use if I were doing this would be some volumetric flasks, a pipette, and a burette. Lets take the simplest example, and I will leave to you to figure out what will work for you. Weigh exactly 0.1 equivalent of KHP and carefully put it into a 1 litre volumetric flask. Fill the flask exactly to the neck mark with distilled water. Now you have a solution of 0.1N KHP. This is your calibration solution. Now prepare your NaOH solution the same way--about 4 g/litre in another volumetric flask. Careful, NaOH will dissolve lots of stuff you don't want to dissolve, such as your eyes and skin and anything wool. You need to know what you're doing in this step. Now, you have a flask of exactly 0.1N acid (the KHP) and approximately 0.1N NaOH. Even if you weigh it precisely, the NaOH picks up so much stuff from the atmosphere that your solutions always need to be calibrated.

Still with me? Now you need a pipette. Put a precise amount of your KHP solution in a flask. Put some of your NaOH solution into a burette. Here you need to know how to titrate. It isn't rocket science, but it does require good technique. Practice! You will need to be pretty good when you get to acetic acid.

You need an indicator--something that changes color when the pH changes. The pH change for these two materials is very fast, and at the "end point" one drop will make the difference in color. Proceed with appropriate caution. So, if you put 10 ml of KHP in your flask, you should expect to see the end point when ABOUT 10 ml of NaOH has been added. If you add precisely 10 ml, your NaOH solution is 0.1N. If you add 11 ml, the solution is obviously weaker than 0.1 N, and would be in fact 0.1 x (10/11) N. Not to belabor the obvious, but your KHP solution has 0.1 equavilents/litre. So, the equivalents in your flask will be 0.1 x ml added/1000. The equivalents of NaOH added will be equal at the end point, so you will know how many equivalents of NaOH you added from your burette. Divide that by the volume, and you have the normality of the NaOH. That is defined as equivalents/litre. Most of work with milliequivalents/millilitre, which is numerically the same.

You won't hit 0.100N. That's fine. You may go through life without ever hitting exactly 0.1000 N NaOH. All you need to know is exactly what it is, and you can calculate everything else.

I hope my ancient memory was reliable and helpful.
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