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Higher Order Derivatives

What is the ...

Question

What is the $[H^{+}]$ in 1M $H_{3} P O_{4}$ if
$H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-} {K a}_{1} = 10^{- 4}$
$H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2} {K a}_{2} = 10^{- 9}$
$H {P O}_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{3 -} {K a}_{3} = 10^{- 13}$

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Solution

$\begin{matrix} H_{3} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-} K a_{1} = 10^{- 4} H_{3} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{- 2} K a_{2} = 10^{- 9} H P O_{4}^{- 2} ⇌ H^{+} + P O_{4}^{3 -} K a_{1} = 10^{- 13} 1 M H_{3} P O_{4} [H_{3} P O_{4}] = 1 M [H^{+}] = ? K O_{4} = \frac{[H^{+}] [H_{3} P O_{4}^{-}]}{H_{3} P O_{4}} K a_{1} = \frac{x . x}{1 - x} \Rightarrow 10^{- 4} = \frac{x^{2}}{1 - x} 10^{- 4} = \frac{x^{2}}{1 - x} \Rightarrow x^{2} = 10^{- 4} - 10^{- 4} x \Rightarrow x^{2} + 10^{- 4} x - 10^{- 4} = 0 x = \frac{10^{- 4} - \sqrt{10^{- 8} - {4.1.10}^{- 4}}}{2} = \frac{10^{- 4} - \sqrt{10^{- 8} - 4 \times 10^{- 4}}}{2} \end{matrix}$

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Similar questions

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

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-}; K_{a_{1}} = 10^{- 3}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2}; K_{a_{2}} = 10^{- 7}

H P O_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{- 3}; K_{a_{3}} = 10^{- 13}

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

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

[H^{+}]

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

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

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

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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 -}

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_{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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