Removing inorganic carbon from sediment

El Siglo

Alright folks,

Bear with me on this, I'm only new to chemistry. I've to do stable isotope mass spec looking for C13 and N15 and C/N ratios. However, I've to prepare my sediment samples for the machine by acid washing with HCL. Now here's where the problems begin, I was told about molarity, %HCL and dilutions etc... only briefly but I don't know how to work out how much HCL, at what percentage and molarity do I need to achieve inorganic carbon removal. I didn't do leaving cert chemistry and so I'm kind of stuck on this bit.

Does anyone know of any good books, papers etc... that would be useful in working this kind of stuff out?

Thanks in advance!

Adamcp898

What is this for? Something like carbon dating?

tony 2 tone

I used to test soil for Total Organic Carbon, I used cold(room Temp) 10%HCl, added drop wise until the sample stops fizing or effervescesing, then 10% HCl heated to about 40 C, then hot DI water to wash any trace of the HCl out.
I was testing for TOC which may not be the same as you, but the method for getting rid of the Inorganic carbon should be the same.
What type of set up are you using?
Here is a paper that may be of help.
http://www.epa.gov/esd/cmb/research/papers/bs116.pdf

El Siglo

I'm using stable isotopes for sea level analysis. Normally you'd do a foram count and classification (i.e. certain forams are present at certain tidal heights, therefore sea level is X). However, some times these aren't present so I'm using sediment and plant material for my analysis, I'm looking for the organic carbon and nitrogen content as that gives me an isotopic variation that is comparable across the inter-tidal zone. I'm using C/N ratios as these differ between the terrestrial and marine zones, and say to some extent what is terrestrial and what is marine and thus, where the high water occurs and this should give a sea level to some extent.
It's just that I'm not sure how to go about preparing the acid to remove the inorganics. I mean, I'll be acid washing my sediment but some times with Geologists for example, they'll take 37% HCL and dilute it into milliq or distilled water at maybe 10ml in 1L and call it 10% HCL, when it's not. I'm probably going to use maybe 10-15 cm3 of sediment for analysis so it's working out how much acid, at what concentration and at what dilution do I need it to be at to remove inorganic material.

El Siglo

Here's where I'm really stuck, I want to have a 10% HCL dilution, now getting a bottle of HCL I'm stuck, to get 10% HCL what do I have to do with regards a 2M (2N) at 2.5 L bech reagent? My head is done in trying to work out how much I need to get a 10% dilution.

Scrappychimow

El Siglo wrote: »

Here's where I'm really stuck, I want to have a 10% HCL dilution, now getting a bottle of HCL I'm stuck, to get 10% HCL what do I have to do with regards a 2M (2N) at 2.5 L bech reagent? My head is done in trying to work out how much I need to get a 10% dilution.

I find this site below really good for chemistry problems
http://www.chemicalforums.com/index.php?board=2.0

Scrappychimow

If you want to get 10% of HCl from the 2.5L of 2N bottle,
Then , I think you need to do this,

If its 2-mole per litre, then a 2.5L bottle has 5-Moles of HCL in it,
so you want 10% HCl, 10% of 1-litre is 0.2 mol

so 0.2mol/5mol= 0.04L x 2.5L=0.1L or 100ml
so I think you need to make up a 1 litre bottle with 100ml of this solution diluted with de-ionized water.

I'm no expert, so could be wrong

El Siglo

Scrappychimow wrote: »

If you want to get 10% of HCl from the 2.5L of 2N bottle,
Then , I think you need to do this,

If its 2-mole per litre, then a 2.5L bottle has 5-Moles of HCL in it,
so you want 10% HCl, 10% of 1-litre is 0.1 mol

so 0.1mol/5mol= 0.02L x 2.5L=0.05L or 50ml
so I think you need to make up a 1 litre bottle with 50ml of this solution diluted with de-ionized water.

I'm no expert, so could be wrong

Fair play, good try so it was. I tried something like that before, but the problem is that you're assuming that the 2N HCl is 100% concentrated. Geologists do it like this a lot but the problem is that usually you usually only get about 38% HCl concentration (I was perhaps a little vague regarding my percentages), so here's what I worked out:

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.

I think that make's sense but I'm unsure, I'll have to wait and see.

Jokesetal

Hi,
Plenty of calculations above, but from a chemists point of view this is very quick and easy.
Concentrated HCl is about 37% (also notice lower case l on Cl).
Therefore to make up a 10% you need to dilute by somewhere in the region of 3.7 to 1 (works out at 27%).
Make up 1 litre of 10% HCl (aq) by adding 270 mL of conc. acid to 730 mL of water slowly!!! Remember acid to water.

37% x 27% = 10%

This is all assuming you're working in the same units (m/v or v/v). Switching between will involve density calculations.

El Siglo

Cheers Jokesetal, that's class so it is.
You wont believe, I went through all this hardship only to find out last Friday that the lab has 10% HCl solution all ready in a squeeze bottle! It was better to work out my dilution in the end it got me thinking a lot.