Chemistry help!?

[Siannon]

Okay so I need a little help with a few questions if someone could explain how to solve these that would be great (:

Prepare 500 mL of 0.5 M Na2EDTA(there's a black dot in the middle here)2H2O
Weigh
out the correct amount of EDTA sodium salt. Show the calculation setup
used to determine amount needed. -----How do I do this?

Add the
EDTA to 400mL of deionized water in a 500mL volumetric flask. Stir with a
stir bar on a magnetic stir plate and measure the pH. Add 10 M NaOH
drop by drop until the pH is 8. Is the EDTA completely dissolved? ----Is
it? We didn't finish this in lab so I don't know how to tell....

And
then there's: Preparation of 1.0 M Tris HCI with two different pH
values. How did you standardize the pH meter for this preparation?
---Another thing we didn't do in lab, can anyone explain how I would do
this?

How would you prepare a 7.5 M solution of NH4OAc?

Using
the 7.5 M solution of NH4OAc and the stock solutions that you prepared
yourself show the calculations you would use to prepare 200mL of the
solution below:
5M NH4OAc
0.5 M Glucose
50mM Na2EDTA(black dot)2H20 -----is this a c1v1=c2v2 formula? if not what formula is it?

If you prepare a 1/40 dilution of 50% solution, what is the final concentration of the solution? ---How do you do this?

A
stock solution of enzyme contains 10mg/ml of enzyme. 100mL of this
stock is diluted with 400 mL of buffer. What is the concentration of the
enzyme in the resulting diluted solution? How much enzyme will be
present in 300μl of the resulting dilution?

Thanks soo much!!!!

[Bromo_DragonFly]

First question :

1 M = 1 Molar = 1mol/liter (note : it means 1 mole of compound per 1 liter of solution. not one liter of solvent.)

so you want to make a 0.5 M concentrated solution which means your final solution should contain 0.5Mole of compound (Na₂EDTA.2H₂O) per each liter of solution.

But notice that you need 500ml of solution which is half a liter. So you need 0.25 mole of compound.

You should be able to calculate the Molar-Mass for your compound using its molecular formula (Molecular formula of EDTA is C₁₀H₁₆N₂O₈), so that you can calculate how many grams of it equals 0.25 Mole.

 

Molar -Mass is germ of 1 mole of a compound. EDTA has a molar-mass of 292.24 g/mol. but that's not your exact compound.

the compound in your question is Na₂EDTA.2H₂O which means it's a salt that has absorbed two water molecules to itself. so its Molar-Mass is increased by 36 g/mole by waters, and by 45.98 by sodium atoms.(and remember it has lost two hydrogen atoms to become a salt.)

because each water molecule has a molar mass of 18 g/mole. and each sodium atom has molar-mass around 22.99

Molar-Mass of your molecule (Na₂EDTA.2H₂O) is 372.24 (that's 22.99 + 22.99 + 292.24 + 18 + 18 -1 -1).

this means each mole of it weighs 372.24 grams. so to get a quarter of mole you should weigh 93.06 grams of it. (that's 372.24 divided by 4)

 

So your answer is 93.06 grams of Na₂EDTA.2H₂O is required to make 500ml of 0.5M solution.

 

Extra Info :

it seems you have not fully understood the concepts of Molarity and Molality.(or at least the practical part.)

so I give you a general info on how to make a solution of 1 M (that's one molar) of a compound.

1. calculate how much compound is required to make that solution. (like how we did above) (this means to know Molar-Mass)
2. weigh the required amount of compound.
3. dissolve it in a less volume of required solvent. (for example in your question you needed to make 500ml, so you should dissolve your compound in something like 200 to 300 ml of water. not 500 at first)
4. add proper volume of solvent to the mixture to get to the desired volume. (it's here that you make your final desired volume, like 500ml in your Question. the reason is : you want the final volume of the solution (which contains both solvent and compound together) to be 500ml)

 

Remember : 1 molal concentration means 1 mole of compound is dissolved in 1 Kg of solvent.

So it's different from Molar concentration. ( 1 molal solution is less concentrated than 1 Molar solution,by assuming solvents and compounds be the same.)

 

sorry don't have time for other today, if you're not in a hurry or something wait to see other answers tomorrow.

Your questions are basic ones in chemistry so highly recommended to master this concepts. so that you don't have to flash-back your books later.