1.025g of an impure sample of a weak acid HX(mol mass = 82) is dissolved in 60ml of water and titrated

Question

with 0.25M NaOH. When half of the acid was neutralized the pH was found to be 5.0 and at the equivalence
point the pH is 9.0. Find the % purity of this sample of HX

Pls help me doing this question

Answer 1

Frankly I think this problem is lacking some information. As an estimate this is what I would do.
pH at the half way point is 5.0 which means the pKa of HX is 5.0 and that makes Ka = 1E-5.
At the eq point you have the hydrolysis of X^- (the anion) as
.....X^- + HOH ==> HX + OH^-
I....y.............0.....0
C...-z.............z.....z
E...y-z............z.....z
The problem tells you that the pH at the eq point is pH = 9. That makes pOH = 5 because pH + pOH = pKw = 14. So (OH^-) = 1E-5.
Kb for X = (Kw/Ka for HX) = (z)(z)/(y-z)
You know Kw, Ka, and z which allows you to calculate y which = (HX) and my estimate of that is 0.1M but you should go through the math and confirm that. That is 0.1 mol/L. You have it in 60 mL. Convert that to mols in the 60 mL, multiply by 82 to find grams in the 1.025 g sample.
Then % HX = (grams HX/1.025)*100 = ? Post your work if you get stuck.
NOTE: The reason I think this is an estimate(i.e., something is missing in the problem) is because when you calculate (HX) at the equivalence point, that is (HX) AFTER the 0.25 M NaOH has been added so the volume at that point is the 60 mL you started with + the amount of 0.25M NaOH added and I don't see any way to estimate how much of the 0.25M NaOH is used. Therefore, I don't think an unqualified answer for % purity can be given. Perhaps another tutor will see something I don't see.

Answer 2

Thank you DrBob222...
Got the answer as 80% purity.