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Revisiting Tautomerism

H 3 P O 4+ H ...

Question

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

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

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

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

The correct option is B
4.675

$H_{2} P O_{4}^{-}$ is amphiprotic which means:

$pH = \frac{p K_{1} + p K_{2}}{2}$

$= \frac{2.15 + 7.2}{2} = 4.675$

Suggest Corrections

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

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}

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}

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_{3} P O_{4}

H_{3} P O_{4} + H_{2} O ⇋ H_{3} O^{+} + H_{2} P O_{4}^{-}

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

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

H_{2} P O_{4}^{-}

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