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First, since NaOH is a base you have to find the pOH first

so you use the equation -> pOH = -log[NaOH]

pOH = -log[NaOH]

= -log[0.0111]

pOH = 1.955

Then you use this equation -> 14 = pH + pOH to find the pH

14 = pH + pOH

pH = 14 - pOH

= 14 - 1.955

pH = 12.045

and that makes it basic

Hope that helped. ^_^

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βˆ™ 11y ago
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βˆ™ 12y ago

At first calculate pOH from [OH-], the substract this value from 14.

So:

pOH = -log[OH-]= -log[0.002614] = -[-3 + log(2.614)] = 2.58 = 2.6 = pOH

Then, from pH + pOH = 14 (at 25 0C) it follows that pH = 14.0 - pOH = 14.0 - 2.6 = 11.4 = pH

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βˆ™ 13y ago

-log(0.001 NaOH)

= 3 subtracted from 14

= 11 pH

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Q: How do you find pH and poh of a 0.002614M NaOH solution?
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