# Question 2109a

Jun 5, 2015

Start with the dissociation of the ethanoic acid, $C {H}_{3} C O O H$, in aqueous solution to give hydronium cations and ethanoate anions, $C {H}_{3} C O {O}^{-}$.

$C {H}_{3} C O O {H}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + C {H}_{3} C O {O}_{\left(a q\right)}^{-}$

The acid dissociation constant, ${K}_{a}$, is equal to

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}$

When in aqueous solution, the ethanoate ion, which is the conjugate base of ethanoic acid, reacts with water to reform ethanoic acid and produce hydroxide ions.

$C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$

The base dissociation constant, ${K}_{b}$, is equal to

${K}_{b} = \frac{\left[C {H}_{3} C O O H\right] \cdot \left[O {H}^{-}\right]}{\left[C {H}_{3} C O {O}^{-}\right]}$

Notice that both equations contain a form of the weak acid - conjugate base ratio. This means that you can use this ratio to determine a relationship between the concentration of hydronium ions, that of hydroxide ions, and the two dissociation constants.

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} \implies \frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} = {K}_{a} / \left(\left[{H}_{3} {O}^{+}\right]\right)$

Since ${K}_{b}$ uses the reciprocal of this ratio, you can write

$\frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{\left[{H}_{3} {O}^{+}\right]}{K} _ a$

Plug this into the equation for ${K}_{b}$ to get

${K}_{b} = \frac{\left[{H}_{3} {O}^{+}\right]}{K} _ a \cdot \left[O {H}^{-}\right] = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[O {H}^{-}\right]}{K} _ a$ " "color(blue)((!))

Now take a look at the equation for the self-ionization of water

$2 {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

The ionization constant of water is known to be

${K}_{W} = \left[{H}_{3} {O}^{+}\right] \cdot \left[O {H}^{-}\right]$

Plug this into equation color(blue)((!))# to get

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

For conjugate acid/base pair, the product of the two dissociation constants is equal to the ionization constant of water.

${K}_{a} \cdot {K}_{b} = {10}^{- 14}$