How do you get pH and pOH given pka or pkb?

1 Answer
anor277
Oct 11, 2016

pH=-log_10[H_3O^+], and pOH=-log_10[HO^-],

Explanation:

In water, under standard conditions, we know that the following equilibrium occurs:

2H_2O rightleftharpoons H_3O^+ + HO^-

As for any equilibrium, we can measure the equilibrium constant, K_w, of this reaction:

K_w=[H_3O^+][HO^-]=10^-14.

Given that K_w=10^-14, and taking -log_10 of both sides, we get the defining relationship:

-log_10K_w = -log_10[H_3O^+]-log_10[HO^-] = -log_(10)10^-14.

Thus 14=pH+pOH, and this is something that you have to commit to memory.

And thus when quoted pK_a for weak acids etc. you have to solve the equilibrium expresssion:

HA+ H_2O rightleftharpoons H_3O^+ +A^-.

There should be many model answers on these boards. Here is a start, and [here is another attempt.](https://socratic.org/questions/how-does-pka-relate-to-acidity)

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