The procedure for total Ca estimation says that a known quantity of sample mixture was dissolved using conc. HCl & diluted to make up to 100 ml with H2O. From this solution, 20 ml were pipetted out and ammonia buffer was added to maintain the pH 9-10. Then it was titrated against std. EDTA solution using EBT as an indicator. The end point is wine red/ pink colour to blue colour.
If we suspect that our sample mixture contains Mg, NaOH/KOH was need to be added to increase the pH above 10. So that all the Mg will be precipitated as Mg(OH)2.
But I found that while using pure CaCO3 sample at pH > 10, there is a precipitation and the filtrate consumes negligible volume of EDTA. It means all the Ca are precipitated as Ca(OH)2. Why?
Hi Nisha
Ca estimated by atomic absorption spectrometry using nitrous acetylene flame
(Marounek et al 2006)
regards
At pH 10-12 Ca2+ forms Ca(OH)2 that is little soluble. To avoid the precipitations is used the buffer NH3- NH4+. In tittration Mg also reacts with edta. To avoid this interference Mg2+ may be complexed before the titration with edta.
To Nisha: you may use 8-oxy-quinoline in ethanol. Dissolve a little amoouint of complexant in ethanol (it gives yellow colour) and add to the solution. Then proceed to titrate calcium with edta as always ( the colour of the indicator may have a little difference) The difference between the data with comnplexant with the total hardness gives the amount of Mg.
To my knowledge Mg(NO3)2 it self is primary standard. However, in EDTA titrations, EBT indicator is used at pH 10 (use buffer solution). Further, details can be had from Vogel's quantitative analysis or any Analytical Chemistry book by Flashka. I feel, addition of EBT it self is a kind of complexation.
I used to do lots of these "total hardness" (Ca +Mg) and Ca alone titrations 50 years ago when working at the state of No Dak's "water lab". The total determination was done with EBT as the indicator and a squirt of roughly pH 10 ammonia/ammonium buffer added to typ. 25 cc of sample. The Ca-only determination was done with murexide indicator (EBT's color doesn't change much with changes in "free" Ca conc. even though it does complex Ca more strongly than does Mg ) & a small squirt of roughly 10 M, carbonate-free, NaOH to "complex" (nominally precipitate) the magnesium. (Mg = the difference between the two titrations.) It's highly unlikely that you will actually precipitate calcium under those conditions - if you are, you're really overdoing the NaOH addition or it contains a good deal of carbonate .
X-ray fluorescence is the best non-destructive technique for total detection of Ca in samples. It requires low energy ~5keV photon source from which photons are made to strike the sample and resulting X-rays are detected and analysed in low energy X-ray spectrometer.
Article Determination of potassium and calcium in vegetables by X-ra...
I agree with Darryl Siemer's comments. Ca is normally titrated at pH 12.5 using the indicator Calcon(TM) after precipitating Mg2+ as Mg(OH)2. You should not see any precipitation of Ca(OH)2 at that pH. Calcium and magnesium can be titrated together at pH 10 with EBT. EBT is not a good indicator for the calcium-only titration at pH 12.5.
Frans method is rather complicated.The quelation method for Mg is rapid and accurate.
My guess is that your NaOH contains enough carbonate to precipitate the Ca.
One way to make carbonate- free NaOH is to saturate water with NaOH pellets & spin down (centrifuge) the solids - Na2CO3 is insoluble in sat. NaOH soln. A few drops of this conc NaOH soln added 10-25 cc of sample ought to get your pH about right.
Re-reading your original post I now realize that you probably reprecipitated Ca as CaCO3 and not as Ca(OH)2. If you start out with CaCO3 dissolved in, say, HCl, you should boil the acid solution for several minutes to remove the CO2 formed by the reaction between CaCO3 and the acid used. If you do'nt, CaCO3 will re-form when you increase the pH. In the presence of Mg, I advise you to increase the pH to 12.5 before titrating Ca. At that point Mg will quantitatively precipitate as Mg(OH)2, as noted in my previous post.
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