Show that for the titration of a weak acid, HA, with NaOH solution, at half-titer, pH = pKa.

Please explain your answer. Thank you.

HA + NaOH ==> NaA + HOH

Ka for HA = (H^+)(A^-)/(HA) and solve for H^+.
(H^+) = Ka*[(HA)/(A^-)
When HA is half neutralized, an equal amount of NaA has been formed, therefore, (HA) = (A^-) at the half-way point of the titration. (You can prove this over and over by starting with ANY volume of HA and adding half that volume of an equimolar NaOH solution.)Therefore, the term (HA)/(A^-) = 1 and
(H^+) = Ka
Take the negative log of both sides
-log(H^+) = -log Ka
pH = pKa.

To show that at half-titer, the pH of a weak acid (HA) titration with NaOH is equal to the pKa, we need to use the equation for the ionization of the weak acid and consider the conditions at the half-way point of the titration.

The reaction between HA and NaOH produces NaA (the conjugate base of HA) and water (HOH) as shown in the equation:
HA + NaOH → NaA + HOH

The equilibrium constant for the ionization of HA is represented by the Ka value, which is given by:
Ka = (H^+)(A^-) / (HA)

To find the concentration of H^+ (hydrogen ions), we can rearrange the equation as follows:
(H^+) = Ka * (HA)/(A^-)

Now, let's focus on the half-way point of the titration. At this point, an equal amount of NaA has been formed as the amount of HA initially present. This implies that the concentrations of HA and A^- are now equal, i.e., (HA) = (A^-).

Substituting (HA) = (A^-) into the equation, we get:
(H^+) = Ka * ((HA)/(A^-)) = Ka * (1)

Since (HA)/(A^-) is equal to 1 at the half-titer, the expression reduces to:
(H^+) = Ka

Taking the negative logarithm (base 10) of both sides, we obtain:
-log(H^+) = -log(Ka)

By definition, -log(H^+) is the pH, and -log(Ka) is the pKa. Hence, we conclude that:
pH = pKa

Therefore, at the half-titer of a weak acid titration, the pH is equal to the pKa of the weak acid.