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Conjugate Acids and Bases

Calculate pH ...

Question

Calculate pH of $1$ $M$ $N a H C O_{3}$ . Given
$H_{2} C O_{3} + H_{2} O ⇌ H C O_{3}^{-} + H_{3} O^{+};$ $p K_{1} = 6.38$
$H C O_{3}^{-} + H_{2} O ⇌ C O_{3}^{2 -} + H_{3} O^{+};$ $p K_{2} = 10.26$

8.73

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8.32

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6.73

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6.32

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Solution

The correct option is B $8.32$
$H C O_{3}^{-}$ is proton acceptor as well as proton donor and therefore it is amphiprotic. For such case, $p H = \frac{p k_{1} + p k_{2}}{2}$
$H C O_{3}^{-}$ undergoes hydrolysis and form alkaline solution.
$H C O_{3}^{-} + H_{2} O ⇌ H_{2} C O_{3} + O H^{-}$
$H C O_{3}^{-}$ can also ionise to form acidic solution.
$H C O_{3}^{-} + H_{2} O ⇌ H_{3} O^{+} + C O_{3}^{2 -}$
pH of the salt is, pH $= \frac{p k_{1} + p k_{2}}{2}$
$= \frac{6.38 + 10.26}{2} = 8.32$

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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_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

H C O_{3}^{-} + H_{2} O ⇌ {C O}_{3}^{2 -} + H_{3} O^{+}

{C O}_{3}^{2 -} + H_{2} O ⇌ {H C O}_{3}^{-} + {O H}^{-}

{C O}_{2} + H_{2} O ⇌ H_{2} {C O}_{3}

{N H}_{3} + H_{2} O ⇌ {N H}_{4} O H

H C l + H_{2} O ⇌ {C l}^{-} + H_{3} O^{+}

N a H C O_{3}

H^{+}

O H^{-} i o n

H C O_{3}^{-} + H_{2} O i o n i z a t i o n ⇋ C O_{3}^{2 -} + H_{2} O

H C O_{3}^{-} + H_{2} O i o n i z a t i o n ⇋ H_{2} C O_{3} + O H^{-}

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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