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)

100ml of 0.1M H_(3)PO_(4) is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution K_(1)=10^(-3)M K_(2)=10^(-8)M K_(3)=10^(-13)M pH at 2nd equivalent point will be :

100ml of 0.1M H_(3)PO_(4) is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution K_(1)=10^(-3)M K_(2)=10^(-8)M K_(3)=10^(-13)M Solubility of compound A(OH)_(2) in final solution will be :

100ml of 0.1M H_(3)PO_(4) is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution K_(1)=10^(-3)M K_(2)=10^(-8)M K_(3)=10^(-13)M pH after HCI was added will be :

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) pH of solution is

Calculate the concentrations of all the species present in 0.1 M H_3PO_4 solution. Given : K_1=7.5xx10^(-3), K_2=6.2xx10^(-8) and K_3=3.6xx10^(-13) .

Calculate the concentration of all species of significant concentrations presents in 0.1 M H_(3)PO_(4) solution. lf K_(1) = 7.5 xx 10^(-3), K_(2) = 6.2 xx 10^(-8), K_(3) = 3.6 xx 10^(-13)

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) Concentration of H_(2)PO_(4)^(-) is