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Arrhenius Acid

K a 1 , K a 2...

Question

$K_{a_{1}}, K_{a_{2}}, K_{a_{3}}$ values for $H_{3} {P O}_{4}$ are $10^{- 3}, 10^{- 8}, 10^{- 12}$ respectively. If $K_{w} (H_{2} O) = 10^{- 14}$ , then
(i) What is dissociation constant of $H {P O}_{4}^{2 -}$ ?
(ii) What is $K_{b}$ of $H {P O}_{4}^{2 -}$
(iii) What is $K_{b}$ of $H_{2} {P O}_{4}^{2 -}$ ?
(iv) What order of $K_{b}$ of ${P O}_{4}^{3 -} (K_{b_{3}}), H {P O}_{4}^{2 -} (K_{b_{2}})$ and $H_{2} {P O}_{4}^{-} (K_{b_{1}})$ ?

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Solution

$H_{3} P O_{4} K a_{1} = 10^{- 3} ⇌ H_{2} P O_{4}^{-}$
$H_{2} P O_{4} K a_{2} = 10^{- 8} ⇌ H P O_{4}^{2 -}$
$H P O_{4}^{2 -} K a_{3} = 10^{- 12} ⇌ P O_{4}^{3 -}$
(i) dissolution cont of $H P O_{4}^{2 -} = 10^{- 8}$
(ii) Kb of $H P O_{4}^{2 -} = 10^{- 6}$ $(∴ K q = 10^{- 8})$
(iii) Kb of $H_{2} P O_{4}^{2 -} = 10^{- 11}$
(iv) $K b_{1} = 10^{- 11}, K b_{2} = 10^{- 6}, K b_{3} = 10^{- 2}$
$K b_{3} > K b_{2} > K b_{1}$

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Similar questions

K_{a_{1}}, K_{a_{2}}

K_{a_{3}}

H_{3} P O_{4}

10^{- 3}, 10^{- 6}

10^{- 12}

K_{w} (H_{2} O) = 10^{- 14}

K_{b}

H P O_{4}^{-}

H_{2} {P O}_{4}^{-}

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

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

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

[H^{+}]

H_{3} P O_{4}

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-} {K a}_{1} = 10^{- 4}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2} {K a}_{2} = 10^{- 9}

H {P O}_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{3 -} {K a}_{3} = 10^{- 13}

[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}

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

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

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

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

H_{3} {P O}_{4} ⇌ 3 H^{+} + {P O}_{4}^{3 -}

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