[" 23."10mL" of "0.1M" tribasic acid "H_(3)A" is titated with "0.1M" NaOH solution."],[[[H_(3)A]],[([at2^(3-)])/((b))=10^(-4)]]

10 mL of 0.1 M tribasic acid H_(3)A is titrated with 0.1 M NaOH solution. What is the ratio of ([H_(3)A])/([A^(3-)]) at 2^(nd) equivalence point ? Given K_(1)=7.5xx10^(-4), K_(2)=10^(-8), K_(3)=10^(-12) .

10 mL of 0.1 M tribasic acid H_(3)A is titrated with 0.1 M NaOH solution. What is the ratio of [[H_(3)A]]/[[A^(3-)]]" at " 2^(nd) equivalence points? [Given : K_(a1)=10^(-3),K_(a2)=10^(-8),K_(a3)=10^(-12)] (a) cong10^(-4) (b) cong10^(+4) (c) cong10^(-7) (d) cong10^(+6)

pH when 100 mL of 0.1 M H_(3)PO_(4) is titrated with 150 mL 0.1 m NaOH solution will be :

________ ml of 0.1 M H_(2)SO_(4) is required to neutralise 50 ml of 0.2 M NaOH solution

How many ml of 1 (M) H_(2)SO_(4) is required neutralise 10 ml of 1 (M) NaOH solution?

……………. ml of 0.1 M H_(2)SO_(4) is required to neutralize 50 ml of 0.2 M NaOH Solution :

Calculate [H^(+)] at equivalent point between titration of 0.1 M, 25 mL of weak acid HA (K_(a(HA)=10^(-5) with 0.05 M NaOH solution: (a) 3xx10^(-9) (b) 1.732xx10^(-9) (c)8 (d)10

20 mL of 0.1M H_(3)BO_(3) solution on complete netralisation requires ….. mL of 0.05M NaOH solution:

Volume of 0.1 M H_2SO_4 solution required to neutralize 10 ml of 0.1 M NaOH solution is :

The volume of 0.1M H_(2)SO_(4) required to neutralise completely 40mL of 0.2M NaOH solution is