pOH of 0.003 M HCl is .

11 + log3

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11 - log3

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7+ log3

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7 - log3

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Solution

The correct option is A 11 + log3
[ $H^{+}$ ]= 3 x $10^{- 3}$

pH = 3 - log3
pOH = 14 - pH
pOH = 11 + log3

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