Question

The salt of an alkali metal gives violet colour in the flame test. Its aqueous solution gives a white precipitate with barium chloride in hydrochloric acid medium. The salt is

• A. K\textsubscript{2}SO\textsubscript{4}
• B. KCl
• C. Na\textsubscript{2}SO\textsubscript{4}
• D. K\textsubscript{2}CO\textsubscript{3}
• E. Li\textsubscript{2}SO\textsubscript{4}

Hint:

Remember these flame colors: $Li$ (Crimson), $Na$ (Yellow), $K$ (Violet), $Ca$ (Brick Red), $Ba$ (Apple Green). Violet is always Potassium!

Answer 1

The Correct Option is A

Concept: This problem uses qualitative analysis to identify the cation and the anion of an unknown salt.
• Cation Identification: Different alkali metals produce characteristic colors in a flame test. Potassium ($K$) gives a distinct violet (lilac) color.
• Anion Identification: The reaction with $BaCl_2$ in $HCl$ is a specific test for the sulfate ion ($SO_4^{2-}$). Barium sulfate ($BaSO_4$) is a white precipitate that is insoluble in dilute $HCl$.

Step 1: Analyze the flame test result. Violet color in a flame test is the hallmark of the Potassium (K\textsuperscript{+}) ion. This eliminates options (C) and (E).

Step 2: Analyze the precipitate test. The formation of a white precipitate with $BaCl_2$ that persists in $HCl$ medium confirms the presence of the Sulfate (SO\textsubscript{4}\textsuperscript{2--}) ion. While carbonates ($CO_3^{2-}$) also form white precipitates with $BaCl_2$, they dissolve in $HCl$ with the evolution of $CO_2$ gas. Combining $K^+$ and $SO_4^{2-}$, the salt is $K_2SO_4$.