# When adding base to a weak acid, does the concentration of "HA" have a net increase after the endpoint?

May 17, 2017

I assume you mean adding strong base to a weak acid.

At the endpoint (i.e. the inflection point in the graph, the equivalence point), ALL of the weak acid $\text{HA}$ has been titrated, and thus converted to its conjugate base ${\text{A}}^{-}$.

So no, the concentration of acid $\text{HA}$ does not have a net increase after the end point... the concentration of base ($\text{NaOH}$ and ${\text{A}}^{-}$, let's say) does, hence the $\text{pH}$ increases.

AFTER THE END POINT, BEFORE LEVEL-OFF

If you bothered to check Le Chatelier's principle at the endpoint, the generated weak conjugate base ${\text{A}}^{-}$ could slightly associate to form a little weak acid $\text{HA}$ back in a weak-base dissociation equilibrium.

(This would be at, say, $\text{25 - 27 mL}$ on the above graph.)

${\text{A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH}}^{-} \left(a q\right)$

Adding more strong base (say, $\text{NaOH}$) neutralizes generated $\text{HA}$ (removing the product), which pushes the weak conjugate base equilibrium towards forming more of the weak acid $\text{HA}$ (the product). This cycle leads to any generated $\text{HA}$ continuously getting neutralized by the incoming $\text{NaOH}$.

Hence, the concentration of weak acid tries to increase but gets neutralized as it gets generated.

May 18, 2017

This is presented as alternate context to the original answer which is 100% correct. Just offering a different point of view.

The increase in wk acid beyond an equivalent reaction mixture of acid and base is 'insignificant' and pH increases should be assumed to be exclusively due to additional $\left(O {H}^{_}\right)$ titrant.

#### Explanation:

Consider the reaction of one equivalent of acid and one equivalent of base. This produces one equivalent of salt + water and is the equivalent point of the titration. If the conjugate base of the salt is strong enough to abstract a proton from water (hydrolysis), then a weak acid is formed along with an excess of hydroxide ions.

The amount of acid formed by the hydrolysis reaction is equal to the excess hydroxide ions. This is such an insignificant amount of acid that it is generally assumed that additional titrant is solely responsible for pH increases without reaction up to the maximum pH of an alkaline system.

Example:
Given: 0.10M HOAc(aq) + 0.10M NaOH(aq) => 0.10M NaOAc(aq) + water
0.10M NaOAc(aq) => 0.10M Na^+(aq) + 0.10M OAc^-"(aq)
$0.10 M N {a}^{+} \left(a q\right) + H O H \left(l\right) \implies N o \left(h y \mathrm{dr} o l y s i s\right) R e a c t i o n$
0.10M OAc^_"(aq) + HOH(aq) => (x)OAc^-(aq) + (x)OH^(-)(aq)
${K}_{a} \left(H O A c\right) = 1.8 x {10}^{- 5}$ $\left[a c i d . i o n i z a t i o n . r e f e r e n c e\right]$
${K}_{b} = {K}_{w} / {K}_{a} = \text{[HOAc][OH]"/"[OAc]} = {\left(x\right)}^{2} / \left(0.10\right) = \frac{{10}^{- 14}}{1.8 x {10}^{- 5}}$
=> $x = \sqrt{\frac{\left(0.10\right) \left({10}^{- 14}\right)}{1.8 x {10}^{- 5}}} M = 7.5 x {10}^{- 6} M = {\left[H A\right]}_{\text{formed}}$

$\left[H A\right] = 7.5 x {10}^{- 6} M$ is such a small quantity, it is generally assumed that this quantity neutralized by additional titrant is insignificant and increases in pH are exclusively due to the additional base up to the maximum of an alkaline system.