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Question

0.1 milli moles of CdSO4 are present in 10 ml acid solution of 0.08 N HCl. Now H2S is passed to precipitate all the Cd+2ions. What would be the pH of solution after filtering off precipitate boiling of H2S and making the solution 100 ml by adding H2O ?

A
pH =2
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B
pH =4
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C
pH = 6
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D
pH = 7
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Solution

The correct option is D pH =2
CdSO4+HCl+H2SCdS+H2SO4
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 HCl) + 2x0.1 (From H2SO4)= 0.1
Total volume= 100 ml
[H+]=1/100=102M
Thus pH=2

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