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Standard XII

Chemistry

Ideal Gas Law

0.1 milli mol...

Question

0.1 milli moles of $C d S O_{4}$ are present in 10 ml acid solution of 0.08 N HCl. Now $H_{2} S$ is passed to precipitate all the $C d^{+ 2}$ ions. What would be the pH of solution after filtering off precipitate boiling of $H_{2} S$ and making the solution 100 ml by adding $H_{2} O$ ?

pH =2

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pH =4

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pH = 6

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pH = 7

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Solution

The correct option is D pH =2
$C d S O_{4} + H C l + H_{2} S \to C d S + H_{2} S O_{4}$
Initial mole 0.1 10x0.08=0.8 0 0 0.8
At Equilibrium 0.1 0.8 0.1 0.1

Milli moles of $H^{+}$ left = 0.8( From $H C l$ ) + 2x0.1 (From $H_{2} S O_{4})$ = 0.1
Total volume= 100 ml
$[H^{+}] = 1 / 100 = 10^{- 2} M$
Thus $p H = 2$

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