`{:("Column - I (Vol. of NaOH added)" ,, "Column- II (pH of resultant solution)") , ("A) 150 mL of NaOH" ,, "P)" pH = p^(ka2)), ("B)" 75 mL of NaOH ,, "Q)" = pH = p^(KaI)), ("C)" 25 mL of NaOH ,, "R)" pH = 7 + [(pKa_(3) + logc)/(2)]) , ("D)" 100 mL of NaOH ,, "S)" pH = (P_(k_(a2)) + p_(K_(a3)))/(2)):}`

Answer

A-R; B-P; C-Q; D-S

Explanation

`{:( H_3PO_4 + ,NaOH to ,NaH_2PO_4 +H_2O ),(0.1 xx 50, 0.1 xx 25 , -),(5 m "mole " , 2.5 m " mole " , 0 ), ( 2.5 ,- , 2.5 m " mole " ):}`
` underset( _) ( 2.5 m " mole ") `
`pH =P^(Ka_1) `
Further addition of 25 ml NaOH , only `NaH_2PO_4` ( 5 m moles ) will be formed
Further addition of 25 ml NaOH ( Total 75 ml) ` {:(NaH_2PO_4+ ,NaOH to , Na_2 HPO_4 + H_2O),( 5 m " moles" , 2.5 m " moles ", 0), ( 2.5 , -, -), ( 2.5 m " moles " , - ,2.5 m " moles "):}`
` therefore pH =p^(ka_2) `
Further addition of 25 ml NaOH , only ` Na_2 HPO_4`
( 5 m moles ) wll be formed
` pH= (P^(Ka_2)+ P^(Ka_3))/( 2) `
Further addition of 25 ml ( Total 125 ml)
` {:( Na_2HPO_4 +, NaOH to, Na_3PO_4+H_2O) ,( 5 m " moles " , 2.5 m " moles " ,0), ( underset(-) (2.5 m " moles ") , - , 2.5 m " moles " ):} `
` therefore pH = pK^(a_3),` Further addition of 25 ml (Total 150 ml) only `Na_3PO_4 ` salt will be formed
` pH =7+ ( p ^(Ka_3))/(2) + ( 1)/(2) log C `