Given
$N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}$ , $K_{b} = 2.22 \times 10^{- 11}$
Thus, the degree of hydrolysis of 0.04 $M$ $N a N O_{2}$ solution is:

1.14 \times 10^{- 2}

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2.36 \times 10^{- 5}

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0.11

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4.86 \times 10^{- 3}

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Solution

The correct option is B $2.36 \times 10^{- 5}$
$N a N O_{2}$ is a salt of a strong base $(N a O H)$ and a weak acid $(H N O_{2})$ . So, hydrolysis reaction will be:
$N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}$

We know degree of hydrolysis is given as
$h = \sqrt{\frac{K_{h}}{C}}$
and $K_{h} = \frac{K_{w}}{K_{a}}$ (For salt of S.B + W.A)
Now, $K_{a} = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{2.22 \times 10^{- 11}} = 4.5 \times 10^{- 4}$
$K_{h} = \frac{10^{- 14}}{4.5 \times 10^{- 4}} = 2.22 \times 10^{- 11}$
and $h = \sqrt{\frac{K_{h}}{C}} = \sqrt{\frac{2.22 \times 10^{- 11}}{0.04}} = 2.36 \times 10^{- 5}$

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Similar questions

N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}

K_{a} (H N O_{2}) = 7.2 \times 10^{- 4}

M

N a N O_{2}

N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}

K_{b} = 2.22 \times 10^{- 11}

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N a N O_{2}

N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}

K_{a} (H N O_{2}) = 7.2 \times 10^{- 4}

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Calculate the hydrolysis constant $(K_{h})$ of the salt containing $N O_{2}$ . Given the

$K_{a}$ for $H N O_{2}$ = $4.5 \times 10^{- 10}$

N O_{2}^{-}

K_{a}

H N O_{2} = 4.5 \times 10^{- 10}

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