Thank you for your answer! I just wasn`t sure if precipitation (precipitate has to be centrifugated) is that what is wanted by the Lanthanium. I will try to avoid this with lower concentrations.

Ca will combine in the flame zone to form Ca-phosphate; so the Ca-signal is a function of P-conc in the solution. By adding La P will preferably combine with P (REE-phosphates) and Ca is released for analysis. However, this is a question of equilibria and I would recommend urgently, to avoid AAS in this case and do the analysis by ICP-OES; it works fine. Have a look not to use oversaturated solutions and use a series of different dilutions to verify the reliability of the results. The effect of oversaturation becomes evident at the nebular tip, which will get choked slowly by the formation of crystals at the tip. You will find this effect also by other solutions tending to oersaturation like solutions of sulphates (AAS nebulisers are generally less delicate, but also there you can find their effect.

Good luck

D. Zachmann

Thank you for your answer, Mr Zachmann. Good idea, I will try it with a series of different dilutions.And dilute even more. The concentration I used till now is right in the middle of the standard solutions that we have. This is why I`ve chosen this.

Sadly we don`t have ICP here so I have to cope with the AAS.

Calcium phosphates are readily soluble in mineral acids, such as hydrochloric, nitric and perchloric. The sample solutions should be acidified with this acid, up to a few% V / V concentration. The lanthanum solution should also be acidified. It is used in a high concentration, usually several grams per liter. Not only phosphates interfere with the calcium signal. In samples of water from rivers, lakes, etc., it is primarily silicon. The lanthanum can be added to the sample solution in a traditional way, i.e. to the solution vessel (also for calibration solutions) or with the help of two capillaries joining in the tee attached to the nebulizer inlet. Greetings. ZJ

Zbigniew's comments are quite correct and useful. When analysing Ca in natural waters you have to add La as a releaser; this is state of the art. It helps to suppress possible formation of Ca-refractories in the flame. However, I forgot to mention: La-chemicals, if not of the highest purity class (extremely expensive) may contain small amounts of Ca. Make sure about this. Otherwise you will bias your results by adding the La-releaser. You can easily find out by doing a comparing AAS analysis of your pure La-solution with a pure Ca reference concentration. So you have to add the very same amount of La to all solutions in order to cancel out the biasing effect. In addition, monitoring a possible clogging of the nebuliser or the change of neb characteristics is done by checking the signal hight of a defined Ca conc in constant intervals.

Good luck

D. Zachmann

Dear Henning,

You will find a thorough description on La and phosphate interaction in:

https://nepis.epa.gov/Exe/ZyNET.exe/9101YSAA.txt?ZyActionD=ZyDocument&Client=EPA&Index=Prior%20to%201976&Docs=&Query=&Time=&EndTime=&SearchMethod=1&TocRestrict=n&Toc=&TocEntry=&QField=&QFieldYear=&QFieldMonth=&QFieldDay=&UseQField=&IntQFieldOp=0&ExtQFieldOp=0&XmlQuery=&File=D%3A%5CZYFILES%5CINDEX%20DATA%5C70THRU75%5CTXT%5C00000027%5C9101YSAA.txt&User=ANONYMOUS&Password=anonymous&SortMethod=h%7C-&MaximumDocuments=1&FuzzyDegree=0&ImageQuality=r75g8/r75g8/x150y150g16/i425&Display=hpfr&DefSeekPage=x&SearchBack=ZyActionL&Back=ZyActionS&BackDesc=Results%20page&MaximumPages=1&ZyEntry=2#

It is an old document. However, it is a study on La and phosphate solubility.

I suppose you always rinse your capillary with high purity water between the sipping of samples. Unless your phosphate is extremely high, this action should avoid choking your capillary when using La.

If needed, centrifugation and/or filtration can be tested with standard solutions with known concentrations of Ca and phosphate, where you can add La. Before sipping for AAS, if centrifugation does not work well, you can use filters similar to the ones that are used in HPLC before injection, for example. If you have time and find it reasonable, you should document your tests with statiscs evaluations.

I would be curious to know your results.

Hello Ademario,

thank you for the paper, I will read it!

The last sampe I prepared (for test measuring) contained 40 mg/L Ca2+ and some mg of inorganic Polyphosphate (K+ as ionisation buffer). When I added Lanthanium I got a precipitate which was centrifuged. The measured Calcium Concentration in the supernatant was 32 mg/L. So either some Ca2+ got lost in the precipitate or was not atomized because of the poylphosphate.

My real samples with unknown Calcium concentration consist of Polyphosphate and Ca2+ Ions in an estimated Ratio Ca/P of 0.15-0.4.

Thanks for your help!

Another way of solving your problem: the addition of Sr. It is a similar solution, but you may not have the same problems with precipitation.

In relation to La: did you test a similar solution only with Ca and without phosphate in parallel for comparison? Did you perform triplicates? Is this loss level (20%) stable along your linear range?

Your Ca concentration is very high! In my equipment, the linear range is 0-5 mg L-1. You should dilute it at least 8 times, just to touch the linear range. Are you diluting your solutions before La addition? I think it could help.

Hello Henning,

it seems the discussion becomes a bit fuzzy and I am no longer sure whether you try to solve an analytical problem (La as a releaser) or you are looking for solubility (activity) products (or both). Anyway, the different subjects require a quite different approach and also, a straight discussion requires some straight background information of problem including devices, sample handling, sample compositions, scope of study etc , which will result in quite different discussion approaches. On the other hand, you see that dealing with Ca, REE, P, refractories, opens to each side a wide field of interesting problems.

Good luck

D. Zachmann

Dieter Wolfram Zachmann .

The answers above certainly contain good and useful idias for you. My idia is that, I go through a mathmatical solution which might be useful for making some differet theoretical suggestions.

Let us assume the Ca is 120ppm, which will make it = 3x10(-3)M. The Ksp of Ca3(PO4)2 = 2.07x10(-33) = 3Ca(2+) + 2PO4(2-). This means that:

[PO4(2-)]2 = 2.07x10(-33)/(3x10(-3))3, [ PO4(2-)] = sq.root of

2.07x10(-33)/9x10(-9). so [PO4(2-)] = 4.8x10(-13)

By using this value, we will find how much La(3+) will be needed to ppt.

LaPO4, from its Ksp which is 3.7x 10(-23)=[La(3+)]x4.8x10(-13), Solving this equation : [La(3+)] = 7.7x10(-11)M

Now, the concentration of LaPO4, will be either 4.8x10(-13)M or 7.7x10(-11)M , according to their availability. According to the calculation above the precipitate of LaPO4, will form. But, I think at such a low concn. the ppt may not appear as a real ppt, but as a masked compound. If a turbidty was seen, at higher concentration than that I have suggested, centrefugation may be necessary.

Note: For the calculation to be clear please copy it on a sheet of paper ai a correct and usual form. Thank You and GOOD LUCK.

Thanks for all your contributions! In the end it was very easy: diluting more really helped. Now the results where really excellent!

I'm glad that dilution worked. Many times the simplest is the wisest.