# Chemistry

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Imagine that you have been asked to prepare 500ml of a 0.25 mol L solution of Sodium Hydroxide, NaOH. Describe how you would prepare this solution. Include any calculations that you would perform.

chemistry - DrBob222, Thursday, October 14, 2010 at 9:00pm
I assume you meant 0.25 moles/L.
M = moles/L
0.25 = moles/0.500 L
moles = 0.125 moles.
Then moles = grams/molar mass
molar mass NaoH = 40 g/mole
grams = 0.125 x 40 = ??

So you want to place ?? g NaoH in a 500 mL volumetric flask, dissolve in some water, then make to the mark with distilled water.

chemistry - Dexter, Friday, October 15, 2010 at 7:34pm
amazing help - this is what I got too however they also want me to explain the steps I would take to prepare this solution. I'm using a graduated cylinder to measure the sodium hydroxide and need to convert the 5 grams to ml. I'm not sure the steps to take. Do I use the molar concentration formula with Vi x 0.125 mol/L = .500L x 0.25 mol/L ? Doing this doesn't seem right to me because I get the answer 1 L. Please let me know. Thank you so much for your help, it really was appreciated and helped me verify my work.

• Chemistry - ,

See my response above.
Now that I read this part, I'm convinced you are NOT familiar with the volumetric flask and perhaps not with NaOH. I'll start over here.
You want to weigh 5.00 grams NaOH (0.125 x 40 g NaOH/mol NaOH = 5.00 g NaOH) on a balance (triple beam or whatever you are using in the lab --not an analytical balance I wouldn't think since NaOH is so corrosive), place the solid NaOH into a 500 mL volumetric flask, add a little water, swirl until all of the NaOH is dissolved, the add water to the mark of the flask (the mark is located on the neck of the flask). Then pop in the stopper (usually these are ground glass fittings), mix the contents thoroughly and you will have 500 mL of a 0.25 M solution of NaOH (if the NaOH were absolutely pure). Technically, no one ever makes NaOH this way because it isn't pure enough as it comes out of the bottle to make it EXACTLY 0.25 but it would be close. In the real world, we weigh the solid NaOH on a triple beam balance, go through the process outlined above, and that makes a solution that is VERY CLOSE to, but not exactly, 0.25. THEN we use an acid with a known molarity (exactly) and titrate the acid against the base and calibrate the NaOH that way.

• Chemistry - ,

I got lost in all the material I'm learning which is why I posted here. Thank you for helping me sort all this out. I was looking at the problem all wrong thinking that the NaOH was a liquid. Very new to chemistry...slowly but surely getting it, thanks for your help, it made a difference!

• Chemistry - ,

You're quite welcome. Come back anytime. We like for you to show what you can do on your on, then explain what you don't understand about the next step.