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pH=-log[H+]

pH-log(1.2x10^-3)

pH=2.92 since the the pH plus the pOH is always equal 14

14-2.92=11.08 so the pOH is 11.08

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11y ago
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11y ago

- log(1.5 M HBr)

= about 0.2 pH

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14y ago

2.8

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14y ago

2.8

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Q: How do you determine the pH of a solution of 1.5 m of HBr?
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