[El Siglo]

Hi folks,

This is my first time posting and I just discovered the site today! I'm carrying out a HCL wash of inter-tidal sediment using 10% HCL. Now here's my problem, I've been trying to work out how to do this dilution, how much bench reagent do I need, at what molarity etc... I came across this document today, and tried working out dilution using 5M (5N) HCL solution (here's more details on the product). What I did was:

% solution = (molarity X molecular weight)/10

%HCL = (5 X 36.46)/10

%HCL = 182.3/10

18.23% HCL Concentration

Then I did according to the formula on page 4 of that document,

10% = (500ml/18.23)

10% = 27.427ml

Therefore, for a solution add 27.427 of bench stock (5M HCL) to 500ml of distilled water.

Now I know this can't be right so any help would be grately appreciated!

[crosemeyer]

You did your second equation incorrectly.

A much simpler way to convert from 18.23% to 10% is to use the following equation:


                    C₁V₁ = C₂V₂

Where C represents Concentration and V is Volume.

So:                (18.23%)(x) = (10%)(500 mL)

                      x = (5000 mL)/(18.23)

                         x = 274 mL

[Borek]

A much simpler way to convert from 18.23% to 10% is to use the following equation:


                    C₁V₁ = C₂V₂

This method is not only simple, it is also wrong. It works for molar concentration, but not for % w/w.

Actually you need 264 mL of acid, taking into account density changes.

[crosemeyer]

Isn't this a w/v % ??

[Borek]

Good point. It wasn't stated and I assumed it is w/w, but 18.23% for 5M seems to confirm w/v.

Still, even if C₁V₁ = C₂V₂ works for w/v, I prefer to treat it as accidental. w/v is poorly defined and easily gets you into absurd numbers. See http://www.chembuddy.com/?left=concentration&right=concentration-follies

[El Siglo]

Folks,

Thanks a million for that help, it was indeed a w/v that I was looking for. That's excellent help so it is. I was trying for ages to figure this out and going nuts. I ended up at one point using this part of the wiki page for HCL (mol/dm³) for 10% and working back to use molarity as a means of working out my dilution for percentage w/v. I got ridiculous numbers for it. Here's what I posted in a Phys-Chem forum back home:

Say the bench stock is 5M (5N) at 2.5L and I need 610ml for washing my samples at 10% dilution.

Moles
Volume (l) = Molarity (M)
10% HCL is equal to 2.87mol/l (this is from a few books, I got it off wikipedia and it was referenced in Perry's Chemical Engineers' Handbook).

Required dilution
10% HCl = 2.87mol/l
The molarity (M) required for this dilution is:

2.87/0.610L = 6.704M

Dilution:

C1 * V1 = C2 * V2

5 * V1 = 6.704 * 0.610

5* V1 = 4.089

V1 = 4.089/5

V1 = 0.817ml of reagent for sediment, in a % m/v solution of 610.817ml.

[Borek]

It can be all easily calculated. Check out this program:

http://www.chembuddy.com/?left=CASC&right=solution_preparation

http://www.chembuddy.com/?left=CASC&right=concentration_conversion

and these lectures:

http://www.chembuddy.com/?left=concentration&right=toc