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 off H2S and making the solution 100 mL by adding H2O is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A)

2

Answer 2

Cd2+ + H2S $\longrightarrow$CdS$\downarrow$ + 2H+

m.moles 0.1 0.2

Total m.moles of H+ in solution after the reaction = 0.2 + 0.8 = 1

$\therefore$ [H+] = $\frac{1}{100}$ = 0.01 M

$\Rightarrow$ pH = 2.