On adding $\ce{BaCl2}$ solution to $\ce{ SO4^{2-}}$ salt aqueous solution $\ce{ BaSO4}$ is formed as a white precipitate, which confirms the presence of the anion.

While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ SO4^{2-}}$ it read:

'Experiment: Aqueous solution of sample + dilute $\ce{ HCl}$ + $\ce{ BaCl2}$ solution

Observation: white ppt'

Why do we need to add $\ce{ HCl}$?

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    $\begingroup$ I should add that in the lab I was asked to add dil HCl and BaCl2 to a "sodium carbonate extract" instead of an aqueous solution of a salt. $\endgroup$ Feb 11 '18 at 11:01

The method consists of slowly adding a dilute solution of barium chloride to a hot solution of the sulphate slightly acidified with hydrochloric acid

It is customary to carry out the precipitation in weakly acid solution in order to prevent the possible formation of the barium salts of such anions as chromate, carbonate, and phosphate, which are insoluble in neutral solutions; moreover, the precipitate so obtained consists of large crystals, and is therefore more easily filtered

Reference: Vogel's Textbook of 'Quantitative Chemical Analysis 11.72 Sulphate: Determination of sulphate as barium sulphate

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    $\begingroup$ Re: "Barium salts of such anions as chromate, carbonate, and phosphate, which are insoluble in neutral solution" You mean to say that the salt sample being tested may have additional ions such as chromate and carbonate, both also with a white ppt. A white ppt in a neutral solution will make it difficult to distinguish whether it was due to sulphate or carbonate ion. But, in acidic solution, it is clear that it must be only sulphate. Is my interpretation correct? $\endgroup$ Feb 11 '18 at 11:36
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    $\begingroup$ Ideally to eliminate false positives which would otherwise occur in the presence of these anions. They however are soluble in presence of dilute mineral acids and(or bases) but are insoluble in neutral solutions $\endgroup$ Feb 11 '18 at 11:40
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    $\begingroup$ Yes, that's what I meant, a false positive. Thanks for confirming! $\endgroup$ Feb 11 '18 at 11:41

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