# Why would CuSO4-EDTA drops be used in Pb titrations?

We use CuSO4-EDTA drops (4) when preparing our lead titration samples for final titration by EDTA. What mechanism would there be for adding this to a titration? What benefit does it produce on Pb titration? We digest galena with nitric and sulfuric, clean up samples with a 5% sulfate wash, and remaining sample is washed into original beaker with 50% ammonium acetate solution (ammonium acetate/acetic acid buffer). Samples are boiled then cooled before final steps. The samples have ~1.5g ascorbic acid, 4 copper-EDTA drops, and 4 xylenol orange drops added to each sample prior to titration with EDTA.

• Galena is the mineral form of lead sulfide. I'd guess that the copper is added to remove any residual sulfide in solution since the Ksp for copper sulfide is $6\times10^{-37}$ which is more insoluble than lead sulfide which has a Ksp of $3\times10^{-28}$ – MaxW Mar 8 at 4:43
• @MaxW and both and extremely insoluble. I suspect non-equilibrium and kinetic effects might dominate here. – Gimelist Mar 9 at 11:18

I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $$\ce{[Cu(EDTA)]^2-}$$ solution to your unknown $$\ce{Pb}$$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($$\ce{Cu^2+}$$) has already complexed with EDTA, it won't interfere with the calculations. Meantime, both $$K_{\mathrm{f}}$$s of $$\ce{[Cu(EDTA)]^2-}$$ and $$\ce{[Pb(EDTA)]^2-}$$ are so closed ($$5\times 10^{18}$$ and $$2\times 10^{18}$$, respectively) that following equilibrium exists: $$\ce{[Cu(EDTA)]^2- + Pb^2+ <=> [Pb(EDTA)]^2- + Cu^2+}$$ Where, $$K_{\mathrm{eq}} = \frac {[\ce{(Pb(EDTA))^2-}][\ce{Cu^2+]}}{[\ce{(Cu(EDTA))^2-}][\ce{Pb^2+]}} = \frac{K_{\mathrm{f}(\ce{(Pb(EDTA))^2-})}}{K_{\mathrm{f}(\ce{(Cu(EDTA))^2-})}} = \frac{2}{5} = 0.4$$

Thus, released $$\ce{Cu^2+}$$ cpmplexed with Xylenol Orange to give stable red color at operating pH (e.g., Figure 1A, Ref.2). My assumption is based on EDTA titration of $$\ce{Pb^2+}$$ and $$\ce{Bi^3+}$$ ion mixtures in the presence of Xylenol Orange as an indicator. In that titration, pH of a test sample solution containing $$\ce{Pb^2+}$$ and $$\ce{Bi^3+}$$ is adjusted to $$3$$ and add the indicator (Xylenol Orange). The test solution becomes red due to the complex formation of $$\ce{Bi^3+}$$ and Xylenol Orange (Figure 1B). When titrate with EDTA, at the first end point, the solution becomes yellow, the color of the free indicator at that pH. To titrate for $$\ce{Pb^2+}$$ at the same solution, it is needed to adjust pH to 5 - 5.5 until the appearance of red color again (indicating the complexation of $$\ce{Pb^2+}$$ with Xylenol Orange). Now, the slution can be titrated with EDTA until the turning of red to yellow. It is important to note that solution of $$\ce{Pb^2+}$$ ions does not become red with Xylenol Orange until pH is above 5. That means, the [$$\ce{Pb^2+}$$ + Xylenol Orange] complex is unstable below pH 5.

Also note that use of ammonium acetate ($$\ce{NH4OAc}$$) as a neutral buffer is misleading. $$\ce{NH4+}$$ and $$\ce{AcO-}$$ are not a conjugate acid/base pair, which means that they do not constitute a buffer at pH 7. However, dissolution of ammonium acetate salt in water results in pH 7, but this pH is highly labile (Ref.1). Ammonium acetate does provide buffering around pH 4.75 (the $$\mathrm{p}K_{\mathrm{a}}$$ of acetic acid) and around pH 9.25 (the $$\mathrm{p}K_{\mathrm{a}}$$ of ammonium). Thus, advantage for using $$\ce{NH4OAc}$$ is a drop from pH 7 to around pH 4.75, which is considerably less dramatic (e.g., addition of $$\pu{1 mM} \: \ce{H+}$$ to a $$\pu{10 mM}$$ neutral pH solution will decrease the pH to 5.8) than doing the same to pure water. Remember, during the titration, with out "real" buffer, pH of the solution changed due to release of $$\ce{H+}$$ ions by complexation.

References:

1. L. Konermann, “Addressing a Common Misconception: Ammonium Acetate as Neutral pH “Buffer” for Native Electrospray Mass Spectrometry,” J. Am. Soc. Mass Spectrom. 2017, 28(9), 1827–1835 (https://doi.org/10.1007/s13361-017-1739-3).
2. Interesting reading: O. J. V. Belleza, A. J. L. Villaraza, “Ion charge density governs selectivity in the formation of metal–Xylenol Orange (M–XO) complexes,” Inorganic Chemistry Communications 2014, 47(1), 87–92 (https://doi.org/10.1016/j.inoche.2014.07.024).
3. Relevant: S. Murakami, “Semi-Xylenol Orange Complexes with Bivalent Metal Ions,” J. inorg, Nucl, Chem. 1981, 43(2), 335–343 (https://doi.org/10.1016/0022-1902(81)90020-8).
• Your chemical equation for the equilibrium depicts the charges on the metal ions wrongly. They should be "2+", not "2-". – Tan Yong Boon Mar 9 at 1:14
• This makes more sense than my guess about sulfide ppts. – MaxW Mar 9 at 16:55
• @Tan Yong Boon: Thanks for the finding. It was a typo and changed accordingly. I appreciate very careful reading. – Mathew Mahindaratne Mar 12 at 15:55
• @MaxW: I glad somebody like it. Sulfide precipitation is a believable suggestion as well. – Mathew Mahindaratne Mar 12 at 15:58
• Thank you for the detailed response. I did some experimentation with Ascorbic and the copper drops on lead nitrate after my post. I did not find any discernable difference between the endpoint characteristics with or without either of these reagents. I need to do further testing with our digested mine concentrate samples to evaluate any variance (e.g. sulfide precipitation). – Acangell Mar 13 at 19:39