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Among the following observations, the correct one that differentiates between $\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ is
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Verified Answer
The correct answer is:
both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ dissolves in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not
$\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ when treated with $\mathrm{BaCl}_{2}$, give white ppt of $BaSO_{3}$ and $BaSO_{4}$ respectively.
$$
\begin{array}{c}
\mathrm{BaCl}_{2}+\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{BaSO}_{3}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphide} \\
\mathrm{BaCl}_{2}+\mathrm{SO}_{4}^{2-} \longrightarrow \begin{array}{c}
\text { } \\
\mathrm{BaSO}_{4}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphate }
\end{array}
\end{array}
$$
Out of these $\mathrm{two}, \mathrm{SO}_{3}^{2-}$ is soluble in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not.
$$
\begin{array}{c}
\mathrm{BaCl}_{2}+\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{BaSO}_{3}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphide} \\
\mathrm{BaCl}_{2}+\mathrm{SO}_{4}^{2-} \longrightarrow \begin{array}{c}
\text { } \\
\mathrm{BaSO}_{4}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphate }
\end{array}
\end{array}
$$
Out of these $\mathrm{two}, \mathrm{SO}_{3}^{2-}$ is soluble in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not.
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