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JEE MainChemistryIonic Equilibrium

50 mL of a 0.2 M weak acid HA is titrated with 0.1 M NaOH . A certain volume V mL of NaOH is added to the acid solution such that the final pH of the mixture becomes 5.60 . The value of V is _ _ _ _ . (Given : p K_a( HA ) = 5.00 , 2 = 0.30 )

Correct answer

80

Step-by-step solution

The reaction between the weak acid and strong base is: HA + NaOH NaA + H ₂ O Initial millimoles of HA = 50 0.2 = 10 mmol Let the volume of NaOH added be V mL . Millimoles of NaOH added = 0.1V mmol Since a buffer is formed, NaOH must be the limiting reagent. Millimoles of salt ( NaA ) formed = 0.1V Millimoles of unreacted acid ( HA ) = 10 - 0.1V Using the Henderson-Hasselbalch equation: pH = p K_a + ( [ Salt ] [ Acid ] ) 5.60 = 5.00 + ( 0.1V 10 - 0.1V ) ( 0.1V 10 - 0.1V ) = 0.60 Since 4 = 2 2 = 2 0.30 = 0.60 , we ha

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