pH and pOH

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752779

Hi guys two things I'm confused about:
frist of all how do you calculate -log(.001) = 3 without using the calculator?

second, why is 3 is not the answer to, why do we have to use pH +pOH =14 and use 14-3 = 11?
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Screen Shot 2016-05-20 at 2.55.23 PM.png
 
Think of .001 as 1 * 10^-3. We know that [-OH] = .001 so we can't calculate the pH directly. But we can calculate the pOH which is equal to -log10([-OH]).

pOH = -log10(1 * 10 ^-3). log10 and 1 * 10 cancel -> pOH = - (-3) = 3.

Using pH + pOH = 14, we get pH = 11.
 
Think of .001 as 1 * 10^-3. We know that [-OH] = .001 so we can't calculate the pH directly. But we can calculate the pOH which is equal to -log10([-OH]).

pOH = -log10(1 * 10 ^-3). log10 and 1 * 10 cancel -> pOH = - (-3) = 3.

Using pH + pOH = 14, we get pH = 11.

Please elaborate on " .001 as 1 * 10^-3"
 
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