Question

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`?

Answer 1

See here...........

Explanation:

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

`pH=pK_a+log_10{(["Asp"^-])/(["HAsp"])}`,

and this represents the equilibrium:

`"HAsp " +" H"_2"O "rightleftharpoons""^(-)"Asp " + " H"_3"O"^+`

Clearly, when `["Asp"^-]=["HAsp"]`, which is the point of half-equivalence in the titration, `pH=pK_a`. Why?