With hot water P $_{2}$ O $_{5}$ gives:
H $_{3}$ PO $_{3}$
H $_{3}$ PO $_{4}$
H $_{3}$ PO $_{2}$
HPO $_{3}$

_{3}

_{4}

_{3}

_{2}

_{3}

_{3}

HPO

_{3}

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In given reactions what are the eq.weight of H₃PO₄?
(1)H₃PO₄ + OH^-= H₂PO ₄ + H₂O
(2)H₃PO₄+ 2OH^- = HPO₄^2-+ H₂O
(3)H₃PO₄ + 3OH^- = PO₄^3- + 3H₂O

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Equivalent weight of

H_{3} P O_{4}

in each of the reaction will be respectively:

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

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H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

Hence, pH of

0.01 M

N a H_{2} {P O}_{4}

is:

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Three reactions involving

H_{2} {P O}_{4}^{-}

are given below:
(i)

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

(ii)

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

(iii)

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

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H_{3} P O_{4} + H_{2} O ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}; p K_{1} = 2.15

H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + H P O_{4}^{- 2}; p K_{2} = 7.20

H e n c e, p H o f 0.01 M N a H_{2} P O_{4} i s

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With hot water P2O5 gives:H3PO3H3PO4H3PO2HPO3