H3PO4-H2O H - 3 O - H -2 PO -4 - pK - 1-2-15 llH2PO - 4- H - 2 O -rightarrow H - 3 O - HPO -2- 4 - pK - 2-7-20 Hence pH of 0-01 M N aH 2 P O 4 is-A- 9-35B- 4-675C- 2-675D- 7-350

The correct option is pH=pK_1+pK_2/2 =2.15+7.2/2=4.675 ...

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Revisiting Tautomerism

H3PO4+H2O H -...

Question

$H_{3} P O_{4} + H_{2} O \to H_{3} O^{+} + H - 2 P O_{4}^{-}; p K_{1} = 2.15 H_{2} P O_{4}^{-} + H_{2} O \to H_{3} O^{+} H P O_{4}^{2 -}; p K_{2} = 7.20$ Hence pH of 0.01 M $N a H_{2} P O_{4}$ is:

9.35

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4.675

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2.675

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7.350

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Solution

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Similar questions

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

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

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

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_{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}

N a H_{2} P O_{4}

\frac{[{H P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}

{p K}_{a} (H_{2} {P O}_{4}^{-}) = 7.1

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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