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Question

0.1 milli mole of CdSO4 are present in 10mL acid solution of 0.08N HCl. Now H2S is passed to precipitate all the Cd2+ ions. The pH of solution after filtering off precipitate, boiling of H2S and making the solution 100mL by adding H2O will be :

A
2
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B
4
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C
8
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D
None of these
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Solution

The correct option is C 2
CdSO4+HCl+H2SCdS+H2SO4
Initially 0.1 0.8 0 0
Finally 0 0.8 0.1 0.1
Milli mole of H+ left in solutio from HCl from H2SO4 respectively =0.8+0.1×2=1.0
Total volume=100mL

[H+]=(1/100)=102M
pH=2

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