Sciencemadness Discussion Board

Preparing base solution for titration - question on molarity

RogueRose - 10-8-2016 at 05:11

If anyone would be so kind as to tell me if the following method is correct for making a proper/accurate 1M solution for titration purposes. I know the math is correct but I'm not sure if the process is correct. If the final volume of a solution, using a 1M mass, is supposed to be 1L (meaning the solution will be heavier than 1g/ml in the case of Sodium Carbonate) then I believe my process is correct. If it is 1L H2O + 1M mass of solvate, then my process is incorrect.

If a 1M solution of Na2CO3 is desired at a quantity of 250ml, the following was examined. The monohydrate was used which has a molar mass of 119.65g/mol

Calculations for monohydrate molar mass
-Na2CO3 anhydrous = 2.54g/cc & 105.98g/mol
-Na2CO3 monohydrate is 2.25g/cc
- Monohydrate/Anhydrous = ratio - 2.25g (monohydrate) /2.54g (anhydrous) = .88583 percent mass of anhydrous:
- 105.98g/mol (anhydrous)/ .88583 (monohydrate Mol conversion) = 119.65g.mol for monohydrate


Making 1/4 of a liter means 1/4 the amount of carbonate
- 119.65g /4 = 29.9125g = amount for 250ml of solution

Volume of 29.9125g of carbonate is subtracted from 250ml
* 29.9125g (amount required) /2.25g (density)= 13.2944ml or cc - displacement
* 250ml - 13.2944ml (carbonate volume) = 236.706ml H2O needed

Mass of 250ml 1M solution:
* 236.706g H2O + 29.9125g Carbonate = 266.6185g
Density of Solution:
* 266.6185g / 250ml = 1.06647g/ml



If 2M solution is desired:

* 59.825g Na2CO3 needed @ volume of 25.589cc = 224.411ml H2O needed

For full liter of 1M
* 119.65g Na2CO3 (molar mass of monohydrate)
* 11.965g / 2.25g/cc = 53.178ml
* 1L solution = 1000ml H2O - 53.178ml Na2CO3 = 946.822ml H2O required
*Density 1.06647g/ml = 946.822 + 119.65 = 1066.472g

Metacelsus - 10-8-2016 at 05:59

Molarity is moles solute per liter solution. Generally, solutions are prepared by dissolving the solute in a somewhat smaller amount of solvent, and then diluting them to the precise volume with more solvent. Your method might work, but it seems overly complicated, and also doesn't account for volume changes due to the effects of a non-ideal solution.

If you wanted to prepare solutions of a specific molality (moles solute per kilogram solvent), then you could just measure the precise amount of solvent and directly add the solute.

RogueRose - 10-8-2016 at 12:03

Quote: Originally posted by Metacelsus  
Molarity is moles solute per liter solution. Generally, solutions are prepared by dissolving the solute in a somewhat smaller amount of solvent, and then diluting them to the precise volume with more solvent. Your method might work, but it seems overly complicated, and also doesn't account for volume changes due to the effects of a non-ideal solution.

If you wanted to prepare solutions of a specific molality (moles solute per kilogram solvent), then you could just measure the precise amount of solvent and directly add the solute.


Thank you! I guess I got confused looking at some solutions (w/v) that weren't water.

So this is what I'm not sure of. You say 1Kg of solvent (in this case water)

So if using the Sodium carbonate and water, then for a 1M solution it would be:

105.98g Anhydrous Na2CO3 + 1L H2O = 1105.98g
105.98g Anhydrous Na2CO3 + 894.02ml H2O = 1KG

The above two lines are why I was confused as to which amount to use for the solvent. I understand the mass of the solute (carbonate) but get messed up on the solvent.

Sorry if I'm being thick, this has been a stumbling block for awhile dealing with this, for me.

**Edit - I finally found this page that actually explains the process very clearly: http://www.chemteam.info/Solutions/Molarity.html
Quote:

Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.

Molarity = 1.00 mol / 1.00 L



So it is clear that the total amount of the solute mass + solvent must equal 1 Liter. So if using something like 100% ethanol as a solvent (which has less mass than H2O @ about.8g/ml) the total mass of the liter would be less than using water, but BOTH would have a total volume of 1L.

Thanks again for the help, this clear sup a lot of things.


[Edited on 10-8-2016 by RogueRose]

[Edited on 10-8-2016 by RogueRose]

DraconicAcid - 10-8-2016 at 12:12

In a volumetric titration, you *always* want to use molarity (with an R), which is moles of solute per litres of solvent. (1 M = 1 mol/L)

If you're doing a titration by mass (which is rarely done), then you would use molality (with an L). (1 m = 1 mol/kg)

To make a 1 M solution of sodium carbonate, you would take the 1 mol of solute (105.98 g) and dissolve it in enough water to make 1 L of solution (usually by dissolving it in a volumetric flask and topping it up to the 1 L mark). Nobody ever carefully measures out the water beforehand.

Procedure

Zandins - 11-8-2016 at 10:28

Standard solutions are typicaly prepared as follows:
1) Weighing out the required quantity of solute;
2) Tranafering to a volumetric flask;
3) Add water to no more than 3/4 of flask's volume;
4)Stopper and shake until dissolved;
5)Wait for the flask to return to room temperature(if applicable) then add water until the mark.

The solution must be taken to room temperature before finalizing its volume to minimize inaccuracy due to thermal expansion.

Hope it helps!

blogfast25 - 11-8-2016 at 11:40

Quote: Originally posted by DraconicAcid  
In a volumetric titration, you *always* want to use molarity (with an R), which is moles of solute per litres of solvent. (1 M = 1 mol/L)



That's wrong and should have been:
Quote:
In a volumetric titration, you *always* want to use molarity (with an R), which is moles of solute per litres of solution. (1 M = 1 mol/L)


When using hydrates, simply bear in mind that 1 mole of hydrate also contains 1 mole of the same, anhydrous reagent. Just adjust the amount to be weighed in accordance with the molar mass of the hydrate.

Example:

1 mol of Na2CO3.10H2O weighs 286.1 g and contains 1 mole of Na2CO3.

1 mol of anh. Na2CO3 weighs 106 g and contains 1 mole of Na2CO3.


zwt - 11-8-2016 at 12:53

Is sodium carbonate a good standard for titrating acids? Doesn't it absorb carbon dioxide from the atmosphere? I suppose you could store it anhydrous in a desiccator with hydroxide, but hydroxide is precious and stock solutions are so much more convenient...
And what's a good standard titrant for bases? Nitric, hydrochloric, and acetic acids are volatile; sulfuric acid is not but is so strong it may be too easy to over-do it and miss the equivalence point. Citric and phosphoric acids are tricky to purify and a magnet for microorganisms. Oxalic acid seems a good candidate, it's cheap and easy to purify; do its solutions degrade at all on standing? (Yes, I've read about KHP, but I'm not satisfied with the availability and pricing I've seen.)

[Edited on 11-8-2016 by zwt]

DraconicAcid - 11-8-2016 at 13:11

Quote: Originally posted by blogfast25  
Quote: Originally posted by DraconicAcid  
In a volumetric titration, you *always* want to use molarity (with an R), which is moles of solute per litres of solvent. (1 M = 1 mol/L)



That's wrong and should have been:
Quote:
In a volumetric titration, you *always* want to use molarity (with an R), which is moles of solute per litres of solution. (1 M = 1 mol/L)

Blast! That's what I meant.

Oxalic acid dihydrate is the best solid standard for acid-base titration, IMHO. Its solutions are stable, it's cheap, and doesn't easily dehydrate. I wouldn't bother using a standard such as sodium carbonate to standardize your acids. I'd use oxalic acid to standardize your sodium hydroxide solution, and use that sodium hydroxide solution to standardize your acids.

[Edited on 11-8-2016 by DraconicAcid]

blogfast25 - 11-8-2016 at 14:25

Quote: Originally posted by DraconicAcid  
I wouldn't bother using a standard such as sodium carbonate to standardize your acids. I'd use oxalic acid to standardize your sodium hydroxide solution, and use that sodium hydroxide solution to standardize your acids.


OA works.

I use twice crystallised soda, then carefully dehydrated, as primary base standard.

NEMO-Chemistry - 13-8-2016 at 13:25

Interesting, i was given the name of an acid to use for practice in titration's, unfortunately i havnt been able to find it for sale.

Oxalic acid is nice and easy and even decent grade is a reasonable price. For base standard i have splashed out on Sodium Carbonate analytical grade and keep in my new desiccator. I couldnt decide on the desiccant so use Calcium chloride in one pot and sodium hydroxide in another.

i did consider sulphuric acid, but i am a bit clumsy so didnt fancy having it splash about.

blogfast25 - 13-8-2016 at 14:42

Quote: Originally posted by NEMO-Chemistry  

i did consider sulphuric acid, but i am a bit clumsy so didnt fancy having it splash about.


Sulphuric acid is no good as a Primary Standard because it is hygroscopic.

Another well known Primary is Potassium Hydrogenphthalate (KHP) but it is not OTC and can be pricey.

NEMO-Chemistry - 13-8-2016 at 14:48

Quote: Originally posted by blogfast25  
Quote: Originally posted by NEMO-Chemistry  

i did consider sulphuric acid, but i am a bit clumsy so didnt fancy having it splash about.


Sulphuric acid is no good as a Primary Standard because it is hygroscopic.

Another well known Primary is Potassium Hydrogenphthalate (KHP) but it is not OTC and can be pricey.


Sorry i was misleading! The sulphuric acid was for the desiccator to keep the carbonate dry, hence the sloshing about when moving the desiccator.

Potassium Hydrogenphthalate is the one mentioned to me in another thread, but as you say its not easy to find for alot of people.

I apologize for the impression i gave that the sulphuric was for use as a standard.