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0.1 millimole of CdSO(4) are present in ...

`0.1` millimole of `CdSO_(4)` are present in 10 mL acid solution of `0.08N HCI`. Now `H_(2)S` is passed to precipitate all the `Cd^(2+)` ions. The pH of the solution after filtering off precipitate, boiling of `H_(2)S` and making the solution 100 mL by adding `H_(2)O`, is:

A

2

B

4

C

6

D

8

Text Solution

Verified by Experts

The correct Answer is:
a
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0.1 millie moles of CdSO_(4) are present in 10ml acid solution of 0.08 N HCI . Now H_(2)S si passed to precipitate all the Cd^(2+) ions. What would be the pH of solution after filtering off percipitate, boilling of H_(2)S and making the solution 100ml by adding H_(2)S ?

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