What! I have no clue with this problem. My teacher didn't explain this well at all! Any help appreciated! Please show steps so maybe I can get what I'm missing!
A 3.381-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 25.0 mL of this solution was titrated with 0.08316-M NaOH. The pH after the addition of 20.89 mL of base was 7.31, and the equivalence point was reached with the addition of 46.01 mL of base.
a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution.
b) What is the molar mass of the acid?
c) What is the pKa of the acid?
2007-11-06 14:50:07 · 2 answers · asked by crazygirl0630 2 in Education & Reference ➔ Homework Help
no i have mr. warren...he's hard!
2007-11-06 14:55:09 · update #1
25.0 mL of the solution of this weak mono-protic acid would contain 3.381g/4 = 0.8453g sample.
(a) From the mole of NaOH used to reach the equivalence point, we can figure out the answer:
4*(20.89mL + 46.01mL)*0.08316M = 22.25 mmole
(b) 3.381g/22.25 mmole = 152.0 g/mol
(c) Let this acid to be represented by HX.
[HX] after mixing but before reaction (to reach pH 7.31):
(20.89+ 46.01)*0.08316/(25.0+20.89) = 0.1212 (M)
[NaOH] after mixing but before reaction (to reach pH 7.31):
20.89*0.08316/(25.0+20.89) = 0.03786 (M)
After reaction HX + NaOH <==> H2O + NaX (with 20.89 mL of base),
[HX] has left: 0.1212M - 0.03786M = 0.0834M
[NaX] has formed: 0.03786M
Hence we have at equilibrium:
HX <==> H+ .... + .... X-
0.0834...10^-7.31...0.03786
Ka = [H+]*[X-]/[HX] = 0.03786*10^-7.31/0.0834 = 2.22x10^-8
pKa = -log(2.22x10^-8) = 7.65
2007-11-10 06:33:12 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋
do you have mr. wedvick?
2007-11-06 22:52:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋