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#?

1 Answer
anor277
Mar 11, 2017

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?

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