PKA VALUES

I understand that the pH at the half equivalence point gives is the pKa value but could someone explain why this point is the pKa value?

HA is a weak acid.
HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
Solve for (H^+)= Ka*(HA)/(A^-)

If we start with, say, 0.1 M HA and start titrating with 0.1 M NaOH, then at the half way point,
1/2 of the 0.1 will have been neutralized leaving exactly 0.05 mols HA remaining. That will have formed exactly 0.05 mols of the salt, NaA, so (HA)=(A^-) so (H^+) = Ka.
Taking the negative log of both sides, leaves
-log(H^+)= - log Ka
pH = pKa.

Neat, huh?

It can also be shown with the Henderson-Hasslebalch equation.
pH = pKa + log (base/(acid)
At the half way point, (base)=(acid) and log 1 = 0, then pH = pKa.

I want to clean up the post a little, mostly for a couple of items omitted.
First, we must take 1 L of the 0.1 M solution of HA. THEN, when we have neutralized exactly half of it, we will have 0.05 mols HA remaining and it will have formed 0.05 mols of the NaA or the anion, A^-. Now the numbers that go into the equation should be molarities, which is mols/L. So If we had 1000 mL (far too much for practical purposes), we will have 0.05 mols HA left and we will have 1000 mL at the beginning + 500 mL of the NaOH that neutralized it and that is a total of 1500 mL. Therefore, the (HA) is 0.05 mols/1.5 L and the (A^-) is 0.05 mols/1.5 L. So (HA)=(A^-), just as my previous post stated and everything works out. I simply didn't want you to confuse molarity (which I started with) with mols (which I ended up with). I left out all of the other logic in between.

Thank you so much that does help a lot. I understood how you went from molarity to mols but thanks for clarifying

  1. 👍
  2. 👎
  3. 👁

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    A 30.0 mL sample of 0.165 molL−1 propanoic acid (Ka=1.3×10−5) is titrated with 0.300 molL−1KOH. Calculate the pH at 0, 5,10, at equivalence point, 1/2 of the equivalence point, 20, 25 mL of added base.

  2. Chemistry

    Explain why the volume of 0.100 M NaOH required to reach the equivalence point in the titration of 25.00 mL of 0.100 M HA is the same regardless of whether HA is a strong or weak acid, yet the pH at the equivalence point is not

  3. chemistry

    The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.220 moles of a monoprotic weak acid (Ka =

  4. Chemistry

    1.) Watch the animation, and observe the titration process using a standard 0.100 M sodium hydroxide solution to titrate 50.0 mL of a 0.100 M hydrochloric acid solution. Identify which of the following statements regarding

  1. Chemistry

    pH indicators change color at their _____. The pH at which the color change happens for a particular indicator molecule depends on its: A. pKa; concentration. B. pKa; Ka or Kb. C. equivalence point; Ka or Kb. D. equivalence point;

  2. Chemistry

    Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. (a) 100.0 mL of 0.14 M HC7H5O2 (Ka= 6.4 multiplied by 10-5) titrated by 0.14 M NaOH halfway point equivalence point (b) 100.0

  3. Chemistry

    35.0 mL of a 0.250 M solution of KOH is titrated with 0.150 M HCl. After 35.0 mL of the HCl has been added, the resultant solution is: A) Acidic and after the equivalence point B) Basic and after the equivalence point C) Neutral

  4. AP Chemistry

    Assume that 30.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of a monoprotic strong acid HX. (a) How many moles of HX have been added at the equivalence point? (b) What is

  1. chemistry

    A 25.00 mL sample of 0.4 M dimethylamine (CH3)2NH is titrated with 0.150 M HCl. What is the pH at the half-equivalence point? Not sure what to do here exactly, I'm bad with the titration problems. I know there are normally a

  2. chemistry

    Which of the following statements is true concerning the titration of a weak base by a solution of hydrochloric acid? Question 9 options: At the equivalence point, the pH is 7. At the equivalence point, there is excess

  3. chemistry

    Why does the equivalence point occur at different pH values for the four titration studied? the four titrations were: 1. HCL with NaOH 2.HC2H3O2 with NaOH 3. HCl with NH4OH 4. HC2H3O2 with NHOH Okay so equivalence point is when

  4. chemistry

    A volume of 100mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the

You can view more similar questions or ask a new question.