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breaking down HCL

Posted: 03 Nov 2006, 10:23
by Dianek
Hi ALL, Can anyone tell how to break hcl down to 0.1,0.01,0.001m??
Can i have the formula please.
Ta Diane :lol: :D :mad: :lol:

//EDITED by adam. Reason: make subject more descriptive.

Posted: 03 Nov 2006, 11:27
by Whspa
Hi Diane,

Do you have the folder called "Manual for School Assistants Working in Science"?
If so, you'll find what you need in Appendix 1.

If not, it goes like this:
starting with 35% HCl (=10M) [concentrated HCl as supplied]
add 100mL of 10M HCl to 900mL water = 1000mL 1M HCl
add 200mL of 10M HCl to 800mL water = 1000mL 2M HCl
add 290mL of 10M HCl to 710mL water = 1000mL 3M HCl
add 390mL of 10M HCl to 610mL water = 1000mL 4M HCl
add 580mL of 10M HCl to 420mL water = 1000mL 1M HCl

To make lower concentrations, start with 1M HCl
add 100mL of 1M HCl to 900mL water = 1000mL 0.1M HCl
add 10mL of 1M HCl to 900mL water = 1000mL 0.01M HCl

I would start with 0.1M HCl to make 0.001M HCl
add 10mL of 0.1M HCl to 900mL water = 1000mL 0.001M HCl

Is that what you wanted to know?

Carol

Posted: 06 Nov 2006, 11:45
by Jazz
Hi Diane,

There is formulae that you can used for any dilution (aqueous)
M1* V1=M2 *V2
Where M 1 is molarity of solution to be diluted
V1 is volume of this solution
M2 is desired molarity
V2 is volume of solution required
To find V1
V1 = M2*V2/M1

Hope this will help

Dilutions

Posted: 06 Nov 2006, 12:30
by Robb
Hi All,

Just to reiterate what method Jazz has offered,

I use the same formula but different terms as explained below:

C[sub]1[/sub] V[sub]1[/sub] = C[sub]2 [/sub]V[sub]2[/sub]

Where;
C[sub]1[/sub] = Initial Concentration (the conc of the solution that is going to be used for the dilutions);

V[sub]1[/sub] = Initial Volume (or the Volume needed to be taken from the Concentrated Stock);

C[sub]2[/sub] = Final Concetration (or the Conc you need);

V[sub]2[/sub] = Final Volume (the volume that the vessel is going to be filled to).

I use the term C as concentration because this formula can be carried over to %v/v OR ppm (Parts per Million) etc. as long as the units stay the same for Concentration and Volume then there is no problem.

You can alter the equation as well to suit the needs of what you are after, for example, working backwards to find the original concentration..

Hope this also helps.

Robb.....