BaCl2 gives white precipitate with an aqueous solution of a salt. The acid radical is either SO2−3 or SO2−4.
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Solution
The acid radical is either SO2−3 or SO2−4. Barium chloride reacts with these radicals to form a white precipitate of either barium sulphite or barium sulphate. BaCl2+SO2−3→BaSO3+2Cl− BaCl2+SO2−4→BaSO4+2Cl−