# Help - spiking sediment with known concentration of sulphide

1. Feb 4, 2017

### Sediment Slayer

I need some assistance. I am truly missing something very fundamental.

Background: I have a sodium sulphide stock solution with the conc. of 0.0429M. I prepare five standards through serial dilution (10,000; 5,000; 1,000; 500; 100uM). Buffer is added in a 1:1 with the standard - the standards are used to calibrate a ion-selective electrode. Once the electrode is calibrated I use it to determine the concentration of sulphide in sediment whereby each sample is prepared with buffer in a 1:1 manner (i.e. 30ml sediment + 30ml buffer). None of this is the issue...

So, here is my problem. I want to spike sediment containing 0uM sulphide with 5,000uM using my stock solution (0.0429M).

Here is my attempt at the calculation:
X-mL = ((desired spike conc.)(vol. of sediment))/(conc. of sodium sulphide stock sol'n)
X-mL = ((5,000uM)(30mL))/(0.0429M) = 3.50ml

When I spike my sediment sample (30mL) with the buffer (30mL) with 3.50mL of the sodium sulphide stock solution the concentration is double (10,000uM) than what it should be - 5,000uM.

If I add 1.75ml of the stock solution to 30mL sediment + 30mL buffer I get the desired answer - 5,000uM.

Should I add 3.50mL of buffer when I add 3.50mL of stock to the 30ml sediment + 30ml buffer or is my calculation incorrect?

Any assistance would be greatly appreciated.

2. Feb 4, 2017

### Staff: Mentor

I have problems following what you do.

Is your sediment a slurry? How do you measure 30 mL? What part of these 30 mL is a solid?

3. Feb 4, 2017

### Sediment Slayer

Hi Borek -

The sediment is collected with minimal pore water. Since sediment is not created equal in terms of size and dimension, the recognized protocol is to standardize the sediment by volume. It is collected in syringes at a volume of 30 ml.

I believe the issue may lie in the fact the buffer is added to the standards in equal volume for calibration but the buffer is not added at 3.50ml when I spike the sediment with 3.50ml of stock solution.

4. Feb 4, 2017

### Staff: Mentor

First thing first - the final concentration of sulfide should be calculated using volume of the solvent, not volume of the solvent plus the solid, as teh latter doesn't dilute the sulfide.

If you get the sulfide concentration twice as large as expected, it could mean your sediment is 50/50 solid and a liquid, so the sulfide is in fact diluted not by 30 mL, but by 15 mL.

I am not saying that's it - it is just a first thing that comes to mind.

5. Feb 4, 2017

### Sediment Slayer

I did test out that theory, Borek. When I added 30ml of buffer to 30ml of sediment the volume was 55ml. So yes, the sediment does contain interstitial spaces which is filled by the buffer. Still doesn't account for the concentration being double. Sigh.

6. Feb 4, 2017

### Staff: Mentor

I don't mean interstitial spaces.

Imagine you have a 15 mL stone (spherical solid to make things easier) in 15 mL of water. Total volume is 30 mL, but when you add 3.5 mL of the 0.0429 M sulfide stock solution final concentration of the sulfide is not 0.00448 M as expected (with the final volume being 33.5 mL) but 0.00812 M (because the sulfide was actually diluted to 18.5 mL).

7. Feb 4, 2017

### Sediment Slayer

Unfortunately my issue is the opposite. I have double the amount of sulphide rather than half than expected.

8. Feb 4, 2017

### Staff: Mentor

Isn't it what happens in the model I have described? The real 0.00812 M is almost twice as large as expected 0.00448 M.