99 The hydrolysis constant for OCl is 3.0 107 then dissociation constant (Ka ) for HOCl is :- (1) 3.33 109 (2) 3.33 107 (3) 3.33 108 (4) 3.33 106

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The percentage of hydrolysis in $0.003 M$ aqueous solution of $N a O C N$ is:

( $K_{a}$ for $H O C N$ = $3.33 \times 10^{- 4} M$ )

0.003 M

N a O C N

K_{a}

H O C N = 3.33 \times 10^{- 4}

0.003 M

N a O C N

(K_{a}

H O C N

3.33 \times 10^{- 4} M)

0.003 M

N a I = O C N

(K_{a}

H O C N

3.33 \times 10^{- 4} M)

0.0 X

n \in N

1.3 + {2.3}^{2} + {3.3}^{3} + . . . + n {.3}^{n} = \frac{(2 n - 1) 3^{n + 1} + 3}{4}

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