# When weak acid "HAsp" is titrated with a strong base and pH is plotted against volume of titrant, at what point in the titration will pH=pK_a?

Mar 11, 2017

#### Explanation:

And thus..........

$p H = p {K}_{a} + {\log}_{10} \left\{\left(\left[\text{Asp"^-])/(["HAsp}\right]\right)\right\}$,

and this represents the equilibrium:

${\text{HAsp " +" H"_2"O "rightleftharpoons""^(-)"Asp " + " H"_3"O}}^{+}$

Clearly, when $\left[\text{Asp"^-]=["HAsp}\right]$, which is the point of half-equivalence in the titration, $p H = p {K}_{a}$. Why?