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x 20 M concentration of H + ion must be maintained in a saturate H 2 S 0.1 M to precipitate C d S but not Z n S , if [ Cd 2 + ⁡ ] = [ Zn 2 + ⁡ ] = 0.1 M initially. K s p ( CdS ⁡ ) = 8 × 1 0 - 27 , K s p Z n S = 1 × 10 - 21 K a H 2 S = 1 × 10 - 21 . Z n S will not precipitate at a concentration of H + greater than x 20 M . The value of x is.

Correct answer

2.00

Step-by-step solution

C d S ⇌ C d 2 + 0.1 + S 2 - s For precipitation of CdS [ Cd 2 + ⁡ ] [ S 2 - ⁡ ] = K sp ( CdS ⁡ ) ⁡ 0.1 × [ S 2 - ⁡ ] = 8 × 1 0 - 27 [ S 2 - ⁡ ] = 8 × 1 0 - 26 For precipitation of ZnS [ Zn 2 + ⁡ ] [ S 2 - ⁡ ] = K sp ( ZnS ⁡ ) ⁡ 0.1 × [ S 2 - ⁡ ] = 1 0 - 21 [ S 2 - ⁡ ] = 1 0 - 20 CdS will start precipitate as the conc of [ S 2 - ⁡ ] = 8 × 1 0 - 26 and ZnS will start precipitate as the conc of [ S 2 - ⁡ ] = 1 0 - 20 so we can raise the concentration of [ S 2 - ⁡ ] upto 1 0 - 20 Now H 2 S 0.1 ( 1 - α ) ≈ 0.1 ⇌ 2 H + 2

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