K$_{W}$, pH, pOH, K$_{b}$ and K$_{a}$

Relate water, acid species and base species in a chemical equation.

$$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$$ (33.1)

What would the equilibrium constant be for this reaction?

K$_{W}$ aka the water dissociation constant:

$$K_{W} = [H^{+}][OH^{-}] = 10^{-14}$$ (33.2)

where $[H^{+}]$ is the concentration of hydrogen in moles and $[OH^{-}]$ is the concentration of the hydroxyl (base) in moles. Note, $[H^{+}]$ is also sometimes represented as $[H_{3}O^{+}]$.

What happens when you take the log of this equation?

$$log ([H^{+}][OH^{-}]) = log (10^{-14})$$ (33.3)

$$log [H^{+}] + log [OH^{-}] = -14$$ (33.4)

or

$$-log [H^{+}] + -log [OH^{-}] = 14$$ (33.5)

$$pH + pOH = 14$$ (33.6)

In Equation 32.5, the negative sign is placed before the $\log$ because it has been agreed that the quantity for $pH$ is positive. The negative sign in front of the $\log$ of a decimal ensures that the answer will be positive. Refer to page in the How to do MCAT Math Without a Calculator chapter for more information on logs.

What is the concentration of $[H^{+}]$ and $[OH^{-}]$ in pure water?

1x10$^{-7}$ for each. Intuitively, if you multiply 1x10$^{-7}$ by 1x10$^{-7}$ you get 1x10$^{-14}$. Or a pH of 7 and a pOH of 7, which add up to 14.

How do you convert a pH or a pOH to a concentration of H$^{+}$ or OH$^{-}$?

Multiply the pH or pOH by negative one, then take the log.

Without a calculator, take the pH or pOH, place a negative in front of it and raise this value above ten. For example, the number of moles of OHÐ in a solution with a pOH of 3 is:

pOH = 3
3 x -1 = -3
10$^{-3}$ = $[OH^{-}]$

Conversely, if the pH is 11 (pH + pOH = 14: therefore, 14 - pOH = pH. In this example we said that pOH was equal to 3, a very basic solution) we find the moles of $H^{+}$ to be:

pH = 11
11 x -1 = -11
10$^{-11}$ = $[H^{+}]$

Note: Although it seems pretty clear, its an easy mistake to make: 10$^{-3}$ is 100 million times more concentrated than 10$^{-11}$ - don't get tripped up by the exponents.

How are pH and pOH related?

Inversely, i.e. if one goes up the other goes down. The convention is to use pH to measure acidity, but don't get tricked by a question that uses pOH. For example, in a very acidic environment with a pH of 2.0, the pOH is equal to 12 - both represent the same environment.

Another thing to keep in mind is that a low pH is equal to a high concentration of H$^{+}$ and therefore a very acidic solution. Similarily, a low pOH is also equal to a high concentration of OH$^{-}$ and therefore is a very basic solution:

Now, if we defined K$_{w}$ above, what about K$_{a}$ and K$_{b}$?

K$_{w}$ is the water dissociation constant, i.e. the equilibrium constant for the dissociation of water. Therefore, K$_{acid}$ and K$_{base}$ are the acid and base dissociation constants:

$HA_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + A^{-}_{(aq)}$

$$K_{Acid} = \frac{[H_{3}O^{+}][A^{}]}{[HA]}$$ (33.7)

$BOH_{(aq)} + H_{2}O_{(l)} \rightleftharpoons B^{+}_{(aq)} + OH^{-}_{(aq)}$

$$K_{Base} = \frac{[B^{+}][OH^{-}]}{[BOH]}$$ (33.8)

What does it mean for a molecule to have either a high K$_{a}$ or K$_{b}$?

The molecule is either a strong acid or a strong base. Since a large K$_{a}$ or K$_{b}$ reflects a molecule that is actively moving in the forward direction in the above equations (creating a large numerator in the equations for the equilibrium constants), it is either actively creating large amounts of H$_{3}$O$^{+}$ or OH$^{-}$.

Note: K$_{a}$ and K$_{b}$ are inversely related to each other.