Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: need help calculating a 1M HCL solution
EmmisonJ
Hazard to Self
**




Posts: 89
Registered: 5-1-2009
Member Is Offline

Mood: No Mood

[*] posted on 27-8-2009 at 11:27
need help calculating a 1M HCL solution


I'm trying to figure out the math to get 225ml of a 1M HCL solution from 33% HCL. I was almost able to get there but can't seem to figure out the rest.

To calculate the 1M solution you use 1 mole of hcl gas per 1L of h2o. You will need to know the starting concentration or molarity of your HCL solution, for example if you are using 33% HCL then you know the density is is 1.16g/ml, that means if you have 1000ml of your 33% HCL solution then it weighs 1160g (1.16*1000=1160). We know that 1000ml of h2o weighs 1000g then that means 160g of that solution must be the HCL. Now that we know how many grams of HCL there is per 1L of your 33% HCL solution, then divide that weight by its molar mass. HCL’s molar mass is 36.46g/mol. So divide 160/36.46 and you get 4.38, that means that your 33% HCL solution is a 4.38M solution.

Here's where I'm lost, probably the easiest part but my mind is running in circles:
You want a 1M solution, then just divide 1M by 4.38M = and get you 0.2283, this means you need to dilute the weight of your 4.38M HCL solution by 22.83%. Now we’ll skip over to our actual desired volumetric amount which is 225ml 1M HCL solution, 22.83% of that weight will need to be your 33% HCL but how do I figure it out from here?
View user's profile View All Posts By User
Paddywhacker
National Hazard
****




Posts: 478
Registered: 28-2-2009
Member Is Offline

Mood: No Mood

[*] posted on 27-8-2009 at 12:23


Molarity is per volume, so you can simplify yout thinking by working in weight / volume exclusively.

If your 33% is weight/weight then convert it to weight/volume. Then working out it's molarity is simple ... from memory it should be about 11 molar. Then work out what volume of that you need for a litre of your required molarity.

If you are weighing it out then back-convert that volume to weight, using the density, and weigh it out and make up to volume with distilled water.
View user's profile View All Posts By User
Magpie
lab constructor
*****




Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline

Mood: Chemistry: the subtle science.

[*] posted on 27-8-2009 at 14:53


molarity x volume = molarity x volume

which is another way of saying: moles = moles.

1. Look up molarity of 33% HCl in a handbook, like CRC
2. Then, using the above equation:

(1M)(225 ml) = (M)(V)

M = molarity of your 33% acid, V = volume of the 33% acid to use.

3. Compute V
4. Dilute V ml of 33% acid to 225 ml with water.




The single most important condition for a successful synthesis is good mixing - Nicodem
View user's profile View All Posts By User
anotheronebitesthedust
Hazard to Others
***




Posts: 189
Registered: 24-6-2007
Member Is Offline

Mood: No Mood

[*] posted on 12-11-2009 at 18:09


The best explanation I found was at the wiki answer website.


http://wiki.answers.com/Q/How_do_you_prepare_a_solution_of_1...
View user's profile View All Posts By User

  Go To Top