Calculate pH of 0-05M potassium hydrogen phthalate- KHC8H4O4- H2C8H4O4-H2O- H3O- -HC8H4O4- - pK1-2-94- HC8H4O4-H2O- H3O- -C8H4O4-2 - pK2-5-44-

PH = (pK1+pK2) / 2 #160; #160; #160; = #160; (2.94 + 5.44)/2 #160; #160; #160; = #160; 4.19 ...

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Types of Redox Reactions

Calculate pH ...

Question

Calculate pH of 0.05M potassium hydrogen phthalate, $K H C_{8} H_{4} O_{4}$ .
$H_{2} C_{8} H_{4} O_{4} + H_{2} O \to H_{3} O^{+} + H C_{8} H_{4} O_{4}^{-} . p K_{1} = 2.94$ .
$H C_{8} H_{4} O_{4}^{-} + H_{2} O \to H_{3} O^{+} + C_{8} H_{4} O_{4}^{- 2} . p K_{2} = 5.44$ .

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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

H e n c e, p H o f 0.01 M N a H_{2} P O_{4} i s

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

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

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

H_{2} O + H_{3} P O_{4}^{-} ⇋ H_{3} O^{+} + H P O_{4}^{2 -}

p K_{1} = 2.15

p K_{2}

N a 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 P O_{4}^{2 -}, p^{k_{2}} = 7.20

p^{H}

N a H_{2} P O_{4}

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