H2O - H3PO-4- H3O- - HPO2-4 Where- pK1-2-15 and pK2 - 7-2Then- pH of 0-01 M NaH2PO4 is-

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pH of a Solution

H2O + H3PO-4⇋...

Question

$H_{2} O + H_{3} P O_{4}^{-} ⇋ H_{3} O^{+} + H P O_{4}^{2 -}$ Where, $p K_{1} = 2.15$ and $p K_{2}$ = 7.2

Then, pH of 0.01 M $N a H_{2} P O_{4}$ is:

9.35

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4.675

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2.675

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7.35

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Solution

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Similar questions

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 P O_{4}^{2 -}; p K_{2} = 7.20

p H

0.01 M N a H_{2} P O_{4}

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

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

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

Hence pH of 0.01 M $N a H_{2} P O_{4}$ is:

K H C_{8} H_{4} O_{4}

H_{2} C_{8} H_{4} O_{4} + H_{2} O \to H_{3} O^{+} + H C_{8} H_{4} O_{4}^{-} . p K_{1} = 2.94

H C_{8} H_{4} O_{4}^{-} + H_{2} O \to H_{3} O^{+} + C_{8} H_{4} O_{4}^{- 2} . p K_{2} = 5.44

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

0.01 M

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

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