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[H(3)PO(4)+H(2)O rightleftharpoons H(3)O...

[H_(3)PO_(4)+H_(2)O rightleftharpoons H_(3)O^(+)+H_(2)PO_(4)^(-);pK_(1)=2.15],[H_(2)PO_(4)^(-)+H_(2)O rightleftharpoons H_(3)O^(+)+HPO_(4)^(-);pK_(2)=7.2],[" Hence "pH" of "0.01MNaH_(2)PO_(4)" is "]

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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

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

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

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)

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

H_(4)underline(P_(2))O_(8)+H_(2)O to H_(3)PO_(4)+H_(2)O_(2)