Ionic EquilibriumHard
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
Options
A.2
B.4
C.6
D.8
Solution
Cd2+ + H2S → CdS↓ + 2H+
m.moles 0.1
0.2
Total m.moles of H+ in solution after the reaction = 0.2 + 0.8 = 1
∴ [H+] =
= 0.01 M ⇒ pH = 2.
m.moles 0.1
0.2
Total m.moles of H+ in solution after the reaction = 0.2 + 0.8 = 1
∴ [H+] =
= 0.01 M ⇒ pH = 2.Create a free account to view solution
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