If Ka of HCN-4- 10-10- then the pH of 2-5- 10-1 molar HCNaq is-

The correct option is D 5.0 [ H ^ + ]=√( 4× 10 ^ 10 × 2.5× 10 ^ 1 ) = 10 ^ 5 ∴ P #160; H = log( H ^ + )= log 10 ^ 5 =5 Here answer ...

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pH of a Solution

If Ka of ...

Question

If $K a$ of $H C N = 4 \times 10^{- 10}$ , then the $p H$ of $2.5 \times 10^{- 1}$ molar $H C N (a q)$ is:

4.2

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4.7

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0.47

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5.0

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Solution

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K_{a}

= 4 \times 10^{- 10}

2.5 \times 10^{- 1}

K_{a}

5.76 \times 10^{- 10}

4.8 \times 10^{- 10}

N H_{4}^{+}

H C N

N H_{4}^{+} (a q .) + C N^{-} (a q .) ⇌ N H_{3} (a q .) + H C N (a q .)

K_{a} = 4.9 \times 10^{- 10}

25^{o}

[H^{+}] = [C N^{-}]

[H^{+}] > [H C N]

H C N (a q)

(K_{a} = 4.8 \times 10^{- 11})

K_{a}

H C N

K_{b} ({C N}^{-}) = 2.5 \times 10^{- 5}

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