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Salt of Weak Acid and Strong Base

Calculate the...

Question

Calculate the $p O H$ of 0.05 M solution of potassium hypochlorite $K O C l$ at $298 K$ . Given that acid dissociation constant of $H O C l$ is $3.5 \times 10^{- 8}$ :
Take $l o g (3.5) = 0.54$

6.92

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5.92

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4.92

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3.92

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Solution

The correct option is D 3.92
$K O C l$ is a salt of weak acid $H O C l$ and strong base $K O H$
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (3.5 \times 10^{- 8}) p K_{a} = (8 - 0.54) = 7.46$
$p H = 7 + \frac{1}{2} (p K_{a} + l o g C) p H = 7 + \frac{1}{2} (7.46 + l o g (0.05)) p H = 7 + \frac{1}{2} (7.46 - 1.30) p H = 7 + 3.08 = 10.08 p O H = 14 - p H = 14 - 10.08 = 3.92$

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Similar questions

p O H

0.05 M

K O C l

298 K

K_{a} (H O C l)

3.5 \times 10^{- 8}

l o g (3.5) = 0.54

1.74 \times 10^{- 5}

0.05 M

p H

p H

0.05 M

8.0 \times 10^{- 9}

p H

N a O C l

K_{a} (H O C l) = 3.5 \times 10^{- 8}

\frac{α_{1}}{α_{2}}

H O C l

C H_{3} C O O H

α_{1}

H O C l

α_{2}

C H_{3} C O O H

K_{a_{1}} (H O C l) = 2 \times 10^{- 7}

K_{a_{1}} (C H_{3} C O O H) = 2 \times 10^{- 5}

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