JEE Main Chemistry question
0.1 millimole of CdSO₄ are present in 10 mL acid solution of 0.08 N HCl. Now H₂S is passed to precipitate all the Cd²⁺ ions. The pH of the solution after filtering off precipitate, boiling of H₂S and making the solution 100 mL by adding H₂O is:
Answer: (A)
Solution
Cd²⁺+H₂S→CdS↓+2H⁺ releases 0.2 mmol extra H⁺; total H⁺=0.8(from HCl)+0.2=1.0 mmol in 100 mL → [H⁺]=0.01 M → pH=2.