Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
(N/A) For testing sulphur,the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. If $H_{2}SO_{4}$ were used,lead acetate itself will react with $H_{2}SO_{4}$ to form a white precipitate of lead sulphate,which will interfere with the test.
$H_{2}SO_{4} + Pb(CH_{3}COO)_{2} \rightarrow PbSO_{4(s)} + 2CH_{3}COOH$
$H_{2}SO_{4}$ (Sulphuric acid) reacts with $Pb(CH_{3}COO)_{2}$ (Lead acetate) to form $PbSO_{4}$ (White precipitate).
$H_{2}SO_{4} + Pb(CH_{3}COO)_{2} \rightarrow PbSO_{4(s)} + 2CH_{3}COOH$
$H_{2}SO_{4}$ (Sulphuric acid) reacts with $Pb(CH_{3}COO)_{2}$ (Lead acetate) to form $PbSO_{4}$ (White precipitate).
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Column-$I$ Column-$II$ $P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow$ $A$. Colourless gas $Q$. $Na_2S + \text{dil. } HCl \rightarrow$ $B$. Brown colour $R$. $NaNO_2 + \text{dil. } HCl \rightarrow$ $C$. Rotten egg smell $S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow$ $D$. Paramagnetic
| Column-$I$ | Column-$II$ |
|---|---|
| $P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow$ | $A$. Colourless gas |
| $Q$. $Na_2S + \text{dil. } HCl \rightarrow$ | $B$. Brown colour |
| $R$. $NaNO_2 + \text{dil. } HCl \rightarrow$ | $C$. Rotten egg smell |
| $S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow$ | $D$. Paramagnetic |
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