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For ortho phosphoric acid, H3PO4(aq)+H2O...

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 :

A

`K_(a_1) gt K_(a_2) lt K_(a_3)`

B

`K_(a_1) lt K_(a_2) lt K_(a_3)`

C

`K_(a_1) gt K_(a_2) gt K_(a_3)`

D

`K_(a_1) lt K_(a_2) gt K_(a_3)`

Text Solution

Verified by Experts

The correct Answer is:
c
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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 ?

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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 :

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 0.1M aqueous NaC solution is : (NaC is sodium salt of acid HC)

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