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${BaCl}_{2} + AcOH + K_{2} {CrO}_{4} ⟶ yellow ppt.$

${BaCO}_{3} (s) + K_{2} G_{2} O_{4} + AcOH ⟶ white ppt.$

${BaCO}_{3} (s) + K_{2} {rO}_{4} + AcOH ⟶ No ppt .$

${SrCO}_{3} (s) + K_{2} {CrO}_{4} + AcOH \to No ppt .$

answer is A, D.

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Detailed Solution

Since, Ba gives yellow ppt. with $K_{2} {CrO}_{4}$ in acetic acid and

${BaC}_{2} O_{4}$ is soluble in hot dilute acetic acid.

${SrCO}_{3}$ react with $K_{2} {CrO}_{4}$ is formed no ppt.

Hence, the correct option is a, d.

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