100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. How much volume of the given NaOH must be added such that a buffer H_2PO_4^(-) // HPO_4^(-2)of maximum capacity is formed.

100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolutio

1.	100 ml 0.1 M H_3PO_4 solution is being titrated with 0.1 M NaOHsolution. The pH of the reaction mixture keeps increasing with addition of NaOH. The successive dissociation constant of H_3PO_4 " are "10^(-3) , 10 ^(-6) and 10^(-14)respectively. How much volume of the given NaOH must be added such that a buffer H_2PO_4^(-) // HPO_4^(-2)of maximum capacity is formed.
Answer» 100 ml 150 ml 200 ml 250 ml Solution :` {:(H_3PO_4 +, NAOH to, NaH_2PO_4, +H_2O ),(10 m " moles " , 10 m " moles " ,0,),(-,-, ,10 m " moles " ):}` ` {:(NaH_2PO_4 +, NaOH to , Na_2HPO_4+ H_2O ),(10 m " moles " , 5 m " moles " ,"-" ),( 5 m " moles " , , ):}` To get BUFFER with maximum buffer capacity HALF of the `NaH_2PO_4` should be consumed. In 1st STEP, To get 10 m moles NaOH, volume should be 100 ml. In 2nd, step , To get 5 m moles NaOH , volume should be 50 ml Total vol = 150 ml