# Question 0b1ae

Mar 18, 2017

Here's why that is the case.

#### Explanation:

You know that pure water undergoes auto-ionization to produce hydronium cations and hydroxide anions according to the following equilibrium reaction

$2 {\text{H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

The equilibrium constant for this equilibrium is defined as

${K}_{c} = \left({\left[\text{H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O}\right]}^{2}\right)$

Now, we don't add the concentration of water to the right side of this expression because the concentration of water in pure water is practically constant.

This implies that we can actually group it with the constant on the left side of the equation to get

overbrace(K_c * ["H"_2"O"]^(2))^(color(blue)(K_W)) = ["H"_3"O"^(+)] * ["OH"^(-)]

The constant term that we have on the left side is called the ion product constant for water

color(blue)(ul(color(black)(K_W= ["H"_3"O"^(+)] * ["OH"^(-)])))" " " "color(darkorange)((1))

Now, let's say that we have a generic weak acid $\text{HA}$. You can pick a weak base instead, the calculations are the same.

As you know, weak acids do not ionize completely in water.

${\text{HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O" _ ((aq))^(+) + "A}}_{\left(a q\right)}^{-}$

Here ${\text{A}}^{-}$ is the conjugate base of the weak acid $\text{HA}$.

So, the equilibrium constant for the above equilibrium will be

${K}_{c} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA"] * ["H"_2"O}\right]\right)$

Once again, the concentration of water is assumed constant, so

$\overbrace{{K}_{c} \cdot \left[\text{H"_2"O"])^(color(blue)(K_a)) = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]}$

This time, the constant term that we have on the left side of the equation is called the acid dissociation constant

color(blue)(ul(color(black)(K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"]))))" " " "color(darkorange)((2))

We also know that ${\text{A}}^{-}$, the conjugate base of the acid, can act as a base in water and ionize to produce

${\text{A"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HA"_ ((aq)) + "OH}}_{\left(a q\right)}^{-}$

The equilibrium constant for this equilibrium looks like this

${K}_{c} = \left(\left[\text{HA"] * ["OH"^(-)])/(["A"^(-)] * ["H"_2"O}\right]\right)$

Once again, this will be equivalent to

$\overbrace{{K}_{c} \cdot \left[{\text{H"_2"O"])^(color(blue)(K_b)) = (["HA"] * ["OH"^(-)])/(["A}}^{-}\right]}$

The constant term that we have on the left side of the equation is called the base dissociation constant

color(blue)(ul(color(black)(K_b = (["HA"] * ["OH"^(-)])/(["A"^(-)]))))" " " "color(darkorange)((3))

Now look what happens when we multiply equations $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$ and $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(3\right)}$

K_a * K_b = (["H"_3"O"^(+)] * color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"])))) * (color(red)(cancel(color(black)(["HA"]))) * ["OH"^(-)])/(color(red)(cancel(color(black)(["A"^(-)]))))#

${K}_{a} \cdot {K}_{b} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right]$

But we know from equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$ that

$\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {K}_{W}$

This means that we have

${K}_{W} = {K}_{a} \cdot {K}_{b}$

which, of course, implies that

${K}_{a} = {K}_{W} / {K}_{b}$