Question

Laboratory Report CHM-153 General Chemisiry Principles il DATE NAME IONIZATION CONSTANT OF A WEAK ACID Trial A Trial B Vol. NaOH to reach equivalence point (ML): Vol. NaOH to reach half-equivalence pt. (mL): -_ pH at half-equivalence pt. (pka): Include your calculations showing how you derived pKfrom K *Remember to include your data sheet and Excel plots (titration plots and first derivative plots).

Answer 1

An Equivalence point in a titration curve is that point where all the acicds present in the solution has completely neutralized by the base added. i.e the exact point of neutralization.

If we add futher base after the equivalance point, then there will be no more acid present to be neutralized. So now the solution will have only base in it. So the pH of the solution will cmpletely due to the base added only.

Now from our table we can see that we have added total 50 mL of NaOH solution. If we observe, we can find untill 33 mL of NaOH added, the pH of the solution is steadily increasing and reached at 6.60

After 1 mL more NaOH was added i.e at 34 mL, the pH suddenly increased to 9.84 and goes on increasing.

Again we know pH > 7 refres to pH of a base.

So here we can say the neutralization point or the Equivalance point has reached at pH = 6.60 and after that since no more acid is present in the solution, the further pH values are due to the extra NaOH added.

So

i- Volume of NaOH to reach Equivalence point = 33 mL

ii- Volume of NaOH to reach half - Equivalence point = 33/2 = 16.5 mL

iii- pH at half equivalence point = pKa = around 4.55

iv- So Ka = 10-pKa = 10-4.55 = 2.8 * 10-5