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Auto Protolysis of Water

Calculate the...

Question

Calculate the equilibrium constants for the reactions with water of $H_{2} P O_{4}^{⊝}, H P O_{4}^{2 -}$ and $P O_{4}^{3 -}$ as base. Comparing the relative values of two equilibrium constants of $H_{2} P O_{4}^{⊝}$ with water, deduce whether solutions of this ion in water are acidic or bases. Deduce whether solutions of $H P O_{4}^{2 -}$ are acidic or bases. Given $K_{1}, K_{2}$ and $K_{3}$ for $H_{3} P O_{4}$ are $7.1 \times 10^{- 3}, 6.3 \times 10^{- 8}$ and $4.5 \times 10^{- 13}$ respectively .

1.4 \times 10^{- 12}, 1.6 \times 10^{- 7}, 2.2 \times 10^{- 2}, H_{2} P O_{4}^{⊝}

is acidic and

H P O_{4}^{2 -}

is basic

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2.2 \times 10^{- 12}, 1.6 \times 10^{- 7}, 1.4 \times 10^{- 2}, H_{2} P O_{4}^{⊝}

is basic and

H P O_{4}^{2 -}

is basic

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2.2 \times 10^{- 12}, 1.6 \times 10^{- 7}, 1.4 \times 10^{- 2}, H_{2} P O_{4}^{⊝}

is acidic and

H P O_{4}^{2 -}

is basic

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1.4 \times 10^{- 12}, 1.6 \times 10^{- 7}, 2.2 \times 10^{- 2}, H_{2} P O_{4}^{⊝}

is basic and

H P O_{4}^{2 -}

is acidic

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Solution

The correct option is B $1.4 \times 10^{- 12}, 1.6 \times 10^{- 7}, 2.2 \times 10^{- 2}, H_{2} P O_{4}^{⊝}$ is acidic and $H P O_{4}^{2 -}$ is basic
i. $H_{2} P O_{4}^{⊝} + H_{2} O ⇌ H_{3} P O_{4} + ⊝ O H$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{7.1 \times 10^{- 3}} = 1.4 \times 10^{- 12}$
ii. $H P O_{4}^{2 -} + H_{2} O ⇌ H_{2} P O_{4}^{⊝} + ⊝ O H$
$K_{h} = \frac{K_{w}}{K_{2}} = \frac{10^{- 14}}{6.3 \times 10^{- 8}} = 1.6 \times 10^{- 7}$
iii. $P O_{4}^{3 -} + H_{2} O ⇌ H P O_{4}^{2 -} + ⊝ O H$
$K_{h} = \frac{K_{w}}{K_{3}} = \frac{10^{- 14}}{4.5 \times 10^{- 13}} = 2.2 \times 10^{- 2}$
$H_{2} P O_{4}^{⊝}$ can react with water to yield $H P O_{4}^{2 -}$ and $H_{3} O^{\oplus}$ with a constant $6.3 \times 10^{- 8}$
Its reaction with water to yield $H_{3} P O_{4}$ , $⊝ O H$ has a much smaller equilibrium constant, $1.4 \times 10^{- 12}$ , and so this ion in water is acidic comparision of the constants for $H P O_{4}^{2 -} (4.5 \times 10^{- 3}$ as an acid, $1.6 \times 10^{- 7}$ as base $)$ leads to the conclusion that this ion in solution is basic.

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Similar questions

H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1}

H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2}

H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}

H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}

H_{3} P O_{4}, H_{3} P O_{4} ⇋ H_{2} P O_{4}^{-} + H^{+} (K_{1})

H_{2} P O_{4}^{-} ⇋ H P O_{4}^{2 -} + H^{+} (K_{2})

H P O_{4}^{2 -} ⇋ P O_{4}^{3 -} + H^{+} (K_{3})

Identify Bronsted acids in the reaction : $H_{2} C O_{3} + H P O_{4}^{2 -} ⇌ H_{2} P O_{4}^{-} + H C O_{3}^{-}$

K_{a}

H P O_{4}^{2 -}

H S O_{3}^{-}

4.8 \times 10^{- 13}

6.3 \times 10^{- 8}

H P O_{4}^{2 -}

H S O_{3}^{-}

P O_{4}^{3 -}

S O_{3}^{2 -}

[H^{+}]

[H_{2} {P O}_{4}]

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

[{P O}_{4}^{3 -}]

0.01 M

H_{3} {P O}_{4}

K_{1} = 7.225 \times 10^{- 3}

K_{2} = 6.8 \times 10^{- 8}

K_{3} = 4.5 \times 10^{- 13}

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