For ortho phosphoric acid, `H_3PO_4(aq)+H_2O(aq) to H_3O^(+)(aq)+H_2PO_4^(-)(aq),K_(a_1)`
`H_2PO_4^-(aq)+H_2O(aq) to H_3O^(+)(aq)+H_2PO_4^(-)(aq),K_(a_2)`
`HPO_4^(2-)(aq)+H_2O(aq) to H_3O^(+)(aq)+PO_3^(4)(aq),K_(a_3)`
The correct order of `K_a` values is :

H_2SO_4 in aqueous medium ionises in two steps H_(2)SO_(4)(aq) + H_(2)O(l) to H_(3)O^(+)(aq) + HSO_(4)^(-) (aq), K_(a1) = x HSO_(4)^(-) (aq) + H_(2)O(l) to H_(3)O^(+) (aq) + SO_(4)^(2-), K_(a2) =y What is relation between x and y

H_(2) SO_(4) in aqueous medium ionises in two steps H_(2) SO_(4) (aq) + H_(2) O (i) to H_(3) O^(+) (aq) + HSO_(4)^(-) (aq) , K_(a_(1)) = x HSO_(4)^(-) (aq) + H_(2) O(l) to H_(3) O^(+) (aq) + SO_(4)^(2-) , K_(a_(2)) = y What is relation between x and y ?

H_(2) SO_(4) in aqueous medium ionises in two steps H_(2) SO_(4) (aq) + H_(2) O (i) to H_(3) O^(+) (aq) + HSO_(4)^(-) (aq) , K_(a_(1)) = x HSO_(4)^(-) (aq) + H_(2) O(l) to H_(3) O^(+) (aq) + SO_(4)^(2-) , K_(a_(2)) = y What is relation between x and y ?

(iii) H_(2)SO_(4(aq)) + NaOH_((aq)) to Na_(2)SO_(4(aq)) + H_(2)O_((I))

To calculate K_(W) of water H_(2)O(l)+H_(2)O_(l)toH_(2)O^(+)(aq)+OH^(-)(aq)

H_2CO_3(aq)+H_2O(l)toHCO_3^-(aq) + H_3O^+(aq) HCO_3^(-)(aq) + H_2O(l)to CO_3^(2-)(aq)+H_3O^(+)(aq) Accoding to the equations above, what is the conjugate base of HCO_3^- ?

The equilibrium equation and K_(a) values for the three acids are given at 25^(@)C : HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5) HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6) HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4) The pH of 0.2M aqueous HA solution is :

In aqueous solution, a weak tribasic acid H_3A ionises in the following three steps: (a) H_3A(aq)+H_2O(l)iffH_3O^(+)(aq)+H_2A^(-)(aq) (b) H_2A^(-)(aq)+H_2O(l)iffH_3O^(+)(aq)+HA^(2-)(aq) (c ) HA^(2-)(aq)+H_2O(l)iffH_3O^(+)(aq)+A^(3-)(aq) If the ionisation constants of these three steps are K_1, K_2 and K_3 respectively, then determine the overall ionisation constant of H_3A .