# Equilibrium constant for the neutralization of weak acid by strong base

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?

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:

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

• wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together. – Avarosa Feb 23 at 7:54
• No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali. – andselisk Feb 23 at 7:58
• Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT! – Avarosa Feb 23 at 18:56