Consider a titration of a 25.00 mL propionic acid solution [K a (C 2 H 5 COOH) =
ID: 1089389 • Letter: C
Question
Consider a titration of a 25.00 mL propionic acid solution [Ka(C2H5COOH) = 1.3 x 10-5] with 0.1107 M solution of sodium hydroxide. The volume of a 22.40 mL of NaOH(aq) was needed to reach the equivalence point. Calculate:
a) the concentration of the propionic acid solution before the titration,
b) the pH of the propionic acid solution before the titration,
c) the pH of the solution at half-equivalence point,
d) the pH of the solution at the equivalence point,
e) the pH of the solution when 0.75 mL of the NaOH has been added PAST the equivalence point,
f) the volume (in mL) of 0.1107 M solution of barium hydroxide that would have to be used to reach the equivalence point if it were used instead of the 0.1107 M NaOH solution?
Remember to SHOW ALL WORK for full marks!
Explanation / Answer
a) The reaction between propionic acid and NaOH is
C2H5COOH + NaOH - - - - - > C2H5COONa + H2O
This is 1:1 reaction
No of mole of NaOH consumed =(0.1107mol/1000ml)×22.40ml=0.002480
0.002480moles of NaOH react with 0.002480moles of C2H5COOH
Volume of propionic acid = 25ml
Therefore,
Concentration of propionic acid = (0.002480mol/25ml)×1000ml = 0.0992M
b) The dissociation of propionic acid is
C2H5COOH <- - - - - - > C2H5COO- + H+
Ka = [C2H5COO-] [H+] /[C2H5COOH] = 1.3×10-5
at equillibrium
[C2H5COOH] = 0.0992 - x
[C2H5COO-] = x
[H+] = x
Therefore,
x2 /(0.0992 - x) = 1.3×10-5
as x value is small, we can assume 0.0992 - x = 0.0992
x2/0.0992 = 1.3×10-5
x2 = 1.29×10-6
x = 1.14×10-3
so,
[H+] = 1.14×10-3M
pH = - log[H+]
= - log(1.14×10-3)
= 2.94
c) At half equivalence point pH =pKa
Ka of propionic acid = 1.3×10-5
pKa = - log(Ka)
= - log(1.3×10-5)
= 4.89
Therefore, at equivalence point
pH = 4.89
d) at equivalence point all the moles of propionic acid is converted in to sodium propionate.
Volume is = 25ml - 22.40ml = 47.40ml
[C2H5COO-] = (0.002480mol/47.40ml)×1000ml = 0.0523M
C2H5COO- + H2O - - - - - > C2H5COOH + OH-
Kb = [C2H5COOH] [OH-] /[C2H5COO-]
Kb = Kw/Ka
where, Kw= 1.00×10-14
Kb = 1.00×10-14/1.3×10-5
= 7.69×10-10
Therefore, at equillibrium
x2/(0.0523 - x) = 7.69×10-10
we can assume, 0.0523-x = 0.0523
x2/0.0523 = 7.69×10-10
x2 = 4.02×10-11
x = 6.34×10-6
So,
[OH-] = 6.34×10-6M
pOH = - log[OH-] = - log(6.34×10-6) = 5.20
pH + pOH = 14
pH = 14 - pOH
= 14 - 5.20
= 8.80
e) No of moles of NaOH added= (0.1107mol/1000ml)×0.75ml = 0.0000830moles
[OH-] = (0.0000830mol/48.15ml)×1000ml =0.001724M
pOH = - log(0.001724) =2.76
pH = 14 - 2.76
=11.24
f) The reaction between Ba(OH)2 and propionic acid is
Ba(OH)2 + 2C2H5COOH - - - - - > Ba(C2H5COO)2 + 2H2O
This reaction is 1:2 molar reaction but reaction
So, half of the volume of consumed in NaOH titration is required
Therefore,
Volume of Ba(OH)2 require = 11.20ml