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{:("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)):} |
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Answer» ` 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 additionof 25 mlNaOH ( 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 additionof 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),` Furtheraddition of 25 ml (Total 150 ml)only`Na_3PO_4 ` salt will be formed ` pH =7+ ( p ^(Ka_3))/(2) + ( 1)/(2)log C ` |
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