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pH the Power of H

H2O + H3PO4⇌ ...

Question

$H_{2} O + H_{3} P O_{4} ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}; p K_{1} = 2.15$
$H_{2} O + H_{2} P O_{4}^{-} ⇌ H_{3} O^{+} + H P O_{4}^{2 -}; p K_{2} = 7.20$
Hence, $p H$ 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 $4.675$
$p H$ of $0.01 M$ of $N a H_{2} P O_{4} = \frac{1}{2} (p K_{1} + p K_{2})$
$= \frac{1}{2} (2.15 + 7.20)$
$= \frac{9.35}{2} = 4.675$

Suggest Corrections

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}

N a H_{2} P O_{4}

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

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

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_{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}^{-}

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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