AIIMS2017ChemistryIonic Equilibrium
K_a for HCN is 5 10⁻¹⁰ at 25^ C . For maintaining a constant pH =9 , the volume of 5 M KCN solution required to be added to 10 ~mL of 2 M HCN solution is
Options
- A4 ~mL
- B2.5 ~mL
- C2 ~mL
- D6.4 ~mL
Correct answer
C. 2 ~mL
Step-by-step solution
pH = p K_a+ [ Salt ] [ Acid ] Let the volume of KCN solution required be V ~mL [ KCN ]= 5 V V+10 and [ HCN ]= 10 2 V+10 Now from eqn. (i), aligned & pH =- (5 10⁻¹⁰ )+ [ 5 V V+10 / 10 2 V+10 ] & 9=- (5 10⁻¹⁰ )+ V 4 aligned On solving, V=1.99 2 ~mL