EmmisonJ - 27-8-2009 at 11:27
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?
Paddywhacker - 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.
Magpie - 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.
anotheronebitesthedust - 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...