Equilibrium constant for the neutralization of weak acid by strong base

Question

Acetic acid has a $K_\mathrm{a}$ of $\pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $\ce{NaOH}$?

Given acetic acid

$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$

$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$

So, if we do $K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$, then we get $K_\mathrm{b} = \pu{5.55e-10}$. How do I use this to find $K_\mathrm{eq}$?

I know how to find $K_\mathrm{eq}$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?

Answer 1

Sodium hydroxide is a strong base and is supposed to be fully dissociated. You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:

$$\ce{HOAc + OH- <=> OAc- + H2O}$$

$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$

Since $[\ce{H2O}] = \text{const}$ (reaction medium), $K'[\ce{H2O}] = K = \text{const}$:

$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$

By multiplying both numerator and denominator by $[\ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_\mathrm{a}$ and ionic product of water $K_\mathrm{w}$:

$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$

For acetic acid:

$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$