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PROFESSOR CIMA: As you know, we define the pH through this water
dissociation. So if K_w is this product, you can take the
log, log base 10 of each side.
Or I can solve for the hydroxide ion concentration. The log of the hydroxide ion concentration
is just log K_w minus log hydronium ion concentration.
If I multiply through by minus 1, and define pH as minus log hydronium ion
concentration, and define pOH as minus log OH-, then you have that pOH equals
minus log K_w minus pH. Or this is pOH equals 14 minus pH, which is
often a very handy tool. Let's just plot it right here, so you can
see. So here we have pOH versus pH.
And you can see there's a linear relationship here with the intercept
being 14 minus log K_w. And there we have, what's called, neutral
pH, when you just have pure water.
If I add some mineral acid that fully dissociates, I'm going to
drive the pH lower. Hydronium ion concentration will be increased.
And I'll have to move right along this curve, because of the water
equilibrium. With a strong acid or a strong base, depending
on which way I'm going, it's fully disassociated.
So the only equilibrium that's controlling the hydrogen ion or
hydroxle ion concentration is just the equilibrium with water.
So that's a pretty handy tool. So that is, when you have a strong acid, you
can calculate the pH very easily when it's added to water, just by assuming
it's fully disassociated. So let's add 0.01 molar HCl.
And you might ask, OK, so what is the hydroxide ion concentration?
Well, since this is a strong acid, it fully disassociates.
And so that means the hydronium ion concentration equals the same
concentration as the chloride ion, because each one of these HCl's that I
put into my solution dissociates. Well, you can use this equation.
You know the pH here is 2. So you could just put it in here.
And you can find out that the pOH is 12. Or you could put in for K_w-- oops.
You know that the hydroxide ion concentration is Kw divided by the
hydronium ion concentration. And this is 1 x 10^ -14 over 0.01.
And you get 1.0 x 10^ -12, so the same answer. The pOH is going to be minus the log of that.
And you should get 12. So this is the pH of 2.
So the pOH should equal 14 minus 2, which equals 12.
So you can do it either way.