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H3PO4+H2OhArrH3O^(+)+H2PO4^(-),pK1 = 2.1...

`H_3PO_4+H_2OhArrH_3O^(+)+H_2PO_4^(-),pK_1 = 2.15`
`H_2PO_4^(-) +H_2OhArrH_3O^(+)+HPO_4^(2-),pK_2 = 7.20`
Hence pH of 0.01M `NaH_2PO_4` is

A

9.35

B

4.675

C

2.675

D

7.35

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H_(3)PO_(4)+H_(2) Leftrightarrow H_(3)O^(+)+H_(2)PO_(4)^(-),pK_(1)=2.15 H_3PO_(4)^(-)+H_(2)O Leftrightarrow H_(3)O^(+)+HPO_(4)^(2-), pK_(2)=7.20 Hence pH of 0.01 M NaH_(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_(4)hArr H^(o+) +H_(2)PO_(4)^(Theta), K_(a_(1)) : H_(2)PO_(4)^(Theta) hArr H^(o+) +HPO_(4)^(2-),' K_(a_(2)) : HPO_(4)^(2-) hArr H^(o+) + PO_(4)^(3-), K_(a_(3)): Mark out the incorrect statements:

Shape of H_3PO_4 is

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?

P_(4)+NaOH to PH_(3) uarr+NaH_(2)PO_(2)

P_(4)+NaOH to PH_(3) uarr+NaH_(2)PO_(2)

Calculate the pH of 0.05 M KHC_(8)H_(4)O_(4) H_(2)C_(8)H_(4)O_(4) + H_(2)O hArr H_(3)O^(o+) + HC_(8)H_(4)O_(4)^(o+) pK_(a_(1)) = 2.94 HC_(8)H_(4)O_(4)^(Theta) + H_(2)O hArr H_(3)O^(o+) + C_(8)H_(4)O_(4)^(2-) pK_(a_(2)) = 5.44

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

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

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