A monoprotic acid (HA) is 1% ionised in its aqueous solution of 0.1 M strength. Its pOH will be:

11

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3

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10

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2

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Solution

The correct option is A $11$
$p H + p O H = 14$

$[H^{+}] = 0.1 \times 0.01 = 0.001 M$

Here, $p H = - l o g [H^{+}] = - l o g 0.1 \times 0.01 = 3$

$p O H = 14 - 3 = 11$

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