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pH (1 Viewer)

xiao1985

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yea... usually, at room temp/ pressure, - log {[h] x [oh]} = 14 or soemthing... i forgot... but to calculate it, calculate pOH den take it away from 14... tis pH...
 

+:: $i[Q]u3 ::+

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[OH-][H+] = 10^-14
i.e. the concentration of hydroxide ions x the concentration of hydrogen ions = 10 to the power of -14.

or as xiao said, u can calculate pOH
pOH = -log [OH-]
and pH = 14 - pOH.
 

+:: $i[Q]u3 ::+

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lol... well i chucked out my calculator ages ago so this has no numbers =P

magnesium hydroxide = Mg(OH)2.
so concentration of hydroxide ions [OH-] = 2 x 4.5 x 10^-3
= 9.0 x 10^-3
Now pOH = - log [OH-]
= - log [9.0 x 10^-3]
now when you've done that... say u get an answer n.

pH = 14 - n.

check ur doing log 10, not natural log. um.. make sure ur signs are correct... u should get an answer greater than 7 yeah?
 

CM_Tutor

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Another approach is to say that pH = 14 + log[OH-], where log is log base 10.
 

blam_babe

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Another approach is to say that pH = 14 + log[OH-], where log is log base 10.
Yeah I think that this is the way I was taught it in class, but we basically did a lesson on how to do the [H+] only, not the [OH-] or when it's basic.

Which method would be better in an exam?? Like the non C1V1=C2V2 one is better than C1V1=C2V2. So should I use the above stated one or:

magnesium hydroxide = Mg(OH)2.
so concentration of hydroxide ions [OH-] = 2 x 4.5 x 10^-3
= 9.0 x 10^-3
Now pOH = - log [OH-]
= - log [9.0 x 10^-3]

????
 

CM_Tutor

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Either is fine, as both are valid - in fact, they are the same method, just done in slightly different ways mathematically. The real 'trick' (if you can call it that) in this question is noting that the formula of magnesium hydroxide is Mg(OH)2, and thus that [OH-] = 2 * [Mg(OH)2].
 

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