1.

In `H_(3)PO_(4),` which of the following is trueA. `K_(a)=K_(a_(1))xxK_(a_(2))xxK_(a_(3))`B. `K_(a_(1))ltK_(a_(2))ltK_(a_(3))`C. `K_(a_(1))gtK_(a_(2))ltK_(a_(3))`D. `K_(a_(1))=K_(a_(2))=K_(a_(3))`

Answer» Correct Answer - A::C
`H_(3)PO_(4)hArrH^(+)+H_(2)PO_(4)^(-)Ka_(1)=([H^(+)][H_(2)PO_(4)^(-)])/([H_(3)PO_(4)])" "...(i)`
`h_(2)PO_(4)^(-)hArrH^(+)+HPO_(4)^(2-)K_(a_(2))=([H^(+)][HPO_(4)^(-2)])/([H_(2)PO_(4)^(-)])" "...(ii)`
`HPO_(4)^(2-)hArrH^(+)+PO_(4)^(3-)K_(a_(3))=([H^(+)][HPO_(4)^(2-)])/([HPO_(4)^(2-)])" "...(iii)`
`H_(3)PO_(4)hArr3H^(+)+P_(4)^(3-)K_(a)=([H^(+)]^(3)[PO_(4)^(3-)])/([H_(3)PO_(4)])" "...(iv)`
Multiplying (i),(ii), (ii) and equate with (iv) `K_(a)=K_(aq)xxK_(a_(2))xxK_(a_(3)),therefore(a)` is correct
`1^(st)` acid dissociation constant is highest because dissociation is maximum in first step, less in `2^(nd)` step and minimum in `3^(rd)` step because it is difficult for a negatively changed ion tolose `H_(+)` ion.
`K_(a_(1))gtK_(a_(2))gtK_(a_(3)), therefore(c)` is correct, (b) and (d) are ruled out.


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