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When the $Na_2CO_3$ is completely neutralized by the $HCl$, the solution will be saturated with $CO_2$, so the pH will be lower than 7 - think of carbonated beverages. Although the $CO_2$ bubbles out, not all of it bubbles out - not even if stirred. If you had stopped the titration at pH = 7, the pH reduction from 11 would have been partly from $HCl$ and ...


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In short: a small quantity of additional water will not alter the quantity (moles) of $\ce{OH-}$ of your lye, or $\ce{H3O+}$ of your acid to be characterized. But it will dilute the intensity of the colour of your indicator used; this then hampering the visual inspection of your analysis. True, the addition of additional water to rinse off the drop from the ...


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At very great dilutions, impurities could have some effect on the end point. For example, even previously-distilled water will absorb some small amount of $\ce{CO2}$ from air, making it slightly more acid. However, if the aspirin is present in reasonable amount, that would have negligible effect. So, as stated by @MaxW, consistency, with a moderate ...


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If I summarize, you want to make an EDTA titration of $\ce{Al^{3+}}$ in a solution containing also $\ce{Cr^{3+}}$ and $\ce{Cr^{6+}}$. The $\ce{Cr^{6+}}$ ions are usually in the form $\ce{CrO4^{2-}}$ and they don't interfere with $\ce{Al^{3+}}$ in EDTA titrations. But they can be eliminated by precipitation. The best thing to do is to oxidize $\ce{Cr^{3+}}$ ...


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Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. ...


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Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions. Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as: $\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$ that implies the further conversion of $\ce{...


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