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Standard XII

Chemistry

Polarity and Dipole Moments

An acid A can...

Question

An acid (A) can oxidise $I^{-}$ to $I_{2}$ . one mole of (A) can neutralise one mole of $C a (O H)_{2}$ in aq solution. One mole of (A) in aqueous solution, on hydrolysis, gives two moles of another acid (B) and one mole of liquid (C), 0.98 g of acid (B) gives 2.33 g of white ppt (D) on reaction with excess of $B a C l_{2}$ . Liquid (C) has oxidising as well as reducing property. (C) can decolourize $K M n O_{4} / H^{+}$ , blackened oil paintings. Compound (A) to (D) are identified as:
(A) : $H_{2} S_{2} O_{B}$ (Marshall's acid), (B) : $H_{2} S O_{4}$ , (C) : $H_{2} O_{2}$ and (D) : $B a S O_{4}$
If true enter 1, else enter 0.

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Solution

The compound (A) is $H_{2} S_{2} O_{8}$ also known as Marshall's acid or peroxodisulphuric acid. The compound (B) is $H_{2} S O_{4}$ .
The compound (C) is $H_{2} O_{2}$ and the compound (D) is $B a S O_{4}$
It can oxidize iodide ion to iodine.
On hydrolysis, it gives two moles of sulphuric acid and on mole of hydrogen peroxide.
$H_{2} S_{2} O_{8} + 2 H_{2} O \to 2 H_{2} S O_{4} + H_{2} O_{2}$
0.98 g (0.01 mole ) of sulphuric acid reacts with barium carbonate to form 2.33 g (0.01 mole) of barium sulphate which is a white ppt (compound D).
$H_{2} S O_{4} + B a C O_{3} \to B a S O_{4} + H_{2} O + C O_{2}$
Hydrogen peroxide is an oxidizing as well as reducing agent.
It decolourizes $K M n O_{4} / H^{+}$ as Mn(VII) is reduced. It also blackens oil paintings.

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