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Three reactions involving H(2)PO(4)^(-) ...

Three reactions involving `H_(2)PO_(4)^(-)` are given below
`I. H_(3)PO_(4)+H_(2)OrarrH_(3)O^(+)+H_(2)PO_(4)^(-)`
`II. H_(2)PO_(4)^(-)+H_(2)OrarrHPO_(4)^(2-)+H_(3)O^(+)`
`III. H_(2)PO_(4)^(-)+OH^(-)rarrH_(3)PO_(4)+O^(2-)`
In which of the above does `H_(2)PO_(4)^(-)` act as an acid?

A

II only

B

I and II

C

III only

D

I only

Text Solution

Verified by Experts

The correct Answer is:
A
Acid is `H^(+)` donor.
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For H_(3)PO_(4) , H_(3)PO_(4) rarr H_(2)PO_(4)^(-)+H^(+)(K_(1)), H_(2)PO_(4)^- rarr HPO_(4)^(-)+H^(+) (K_2), HPO_(4)^(2-) rarr PO_(4)^(3-) + H^(+) (K_(3)) then

The conjugate base of H_(2)PO_(4)^(-) is :

The conjugate base of H_(2)PO_(4)^(-) is :

For, H_(3)PO_(4)+H_(2)OhArrH_(3)O^(+)+H_(2)PO_(4)^(-), K_(a_(1)) H_(2)PO_(4)+H_(2)OhArrH_(3)O^(+)+HPO_(4)^(2-), K_(a_(2)) HPO_(4)^(2-)+H_(2)OhArrH_(3)O^(+)+PO_(4)^(3-), K_(a_(3)) The correct order of K_(a) values is:

H_(3)PO_(2)+AgNO_(2) to Ag darr+H_(3)PO_(4)+NO

H_(3)PO_(2)+AgNO_(2) to Ag darr+H_(3)PO_(4)+NO

H_(3)underline(P)O_(5)+H_(2)O to H_(3)PO_(4)+H_(2)O_(2)

H_(3)underline(P)O_(5)+H_(2)O to H_(3)PO_(4)+H_(2)O_(2)

Equivalent weight of H_(3)PO_(4) in each of the reaction will be respectively - H_(3)+OH^(-)rarrH_(2)PO_(4)+H_(2) H_(3)PO_(4)+2OH^(-)rarrHPO_(4)^(2-)+2H_(2)O H_(3)PO_(4)+2OH^(-)rarrPO_(4)^(3-)+3H_(2)O

underline(P_(4))O_(10)+H_(2)O to H_(3)PO_(4)

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