I. If you add 0.439 g of benzoic acid to a 100 ml volumetric flask and then add
ID: 1033521 • Letter: I
Question
I. If you add 0.439 g of benzoic acid to a 100 ml volumetric flask and then add enough water to bring the total volume to 100.00 ml, how many ml of 0.2759 M NaOH solution will be required to completely neutralize the benzoic acid? At the neutralization point, will the solution be somewhat acidic, neutral, or somewhat basic? 2. If you redo the experiment in question 1 using a 50 ml volumetric flask instead of a 100 ml volumetric flask (but the same amount of benzoic acid), how many ml of 0.2759 M NaOH solution will be required to completely neutralize the benzoic acid? 3. If you accidentally overshoot the neutralization point in the experiment described in question 1 and add 0.10 ml too much of the NaOH solution, what will be the final solution pH? 4. Considering your answers to questions 1 and 2 carefully, which of the stock pH indicators (methyl orange, thymol blue, or phenolphthalein) do you want to use for your titrations with the indicator dye? Explain your reasoning. 5. You need to determine both the molar mass and the pK, of your known acid. On which of these experiments will you use the pH meter? (Remember, you can only use the pH meter on one experiment.) Please explain your decision.Explanation / Answer
5) pH meter is used to find out pKa
pKa = - log Ka
where, Ka = acid dissociation constant of an acid
if HA is considered as an acid, which dissociates as,
HA = H+ + A- - ------------------------------ (1)
Ka = [H+][A-] / [HA] ------------------------------- (2 )
from equation (2),
Ka is directly proportional to [H+]
from the reading of the pH meter, H+ ion concentration can be found out.
thus, from equation (1),
[H+] = [A-] and from this data Ka can be found out .
HA = H+ + A-
Initial- X M 0 0
Change- -Y +Y +Y
Concentration X-Y Y Y
Ka = [H+][A-] / [HA] = Y .Y / (X-Y)