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

Peyton Adams

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

Thus,...

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

and this illustrates the state of balance:

#"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?

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Mateo Aguilar

At the halfway point of the titration, when the moles of strong base added equal the moles of weak acid initially present, the pH will be equal to the pKa of the weak acid. This is because at this point, the solution contains equal concentrations of the weak acid and its conjugate base, resulting in a pH equal to the pKa of the weak acid.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.