${H P O_{4}}^{2 -} + H_{2} O ⇌ H_{3} O^{+} + {P O_{4}}^{3 -}$
In the equation above, which compounds are Bronsted bases?

{H P O_{4}}^{3 -}

and

H_{2} O

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H_{2} O

and

{P O_{4}}^{3 -}

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H_{3} O^{+}

and

H_{2} O

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

and

{P O_{4}}^{3 -}

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Solution

The correct option is B $H_{2} O$ and ${P O_{4}}^{3 -}$
A bronsted base is a species that accepts a proton in aqueous solutions.
$H P O_{4}^{2 -} + H_{2} O ⇋ H_{3} O^{+} + P O_{4}^{3 -}$
In the above equilibrium in forward reaction.
$H_{2} O$ accepts a proton to give $H_{3} O^{+}$ .
In backward reaction, $P O_{4}^{3 -}$ accepts a proton to give $H P O_{3}^{2 -}$ .
Hence, $H_{2} O$ and $P O_{4}^{3 -}$ act as Bronsted bases in the given equation.

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Similar questions

H C_{2} O_{4}^{-} + P O_{4}^{3 -} ⇌ H P O_{4}^{2 -} + C_{2} O_{4}^{2 -}

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}

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

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

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

H F + H_{3} O^{+} \to H_{2} F^{+} + H_{2} O

P O_{4}^{3 -} + H_{2} O (l) \to H P O_{4}^{2 -} (a q) + O H^{-} (a q)

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