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Question

pOH of 0.003 M HCl is .

A
11 + log3
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B
11 - log3
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C
7+ log3
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D
7 - log3
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Solution

The correct option is A 11 + log3
[H+]= 3 x103

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

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