Acetic acid is a weak acid with a pKa of 4.75. Set up a titration using 20.00 mL
ID: 1004897 • Letter: A
Question
Acetic acid is a weak acid with a pKa of 4.75. Set up a titration using 20.00 mL of acetic acid solution to 20mL strong base NaOH.
Concentration of NaOH soln: 9.878x10^-2 mol/L
Initial buret reading (mL): 0.00
Buret reading at equivalence pt (mL): 21.51
Calculate the following:
a.) Moles of NaOH added at equivalence pt
b.) Moles of acetic acid in sample
c.) Acetic acid concentration (mol/L)
d.) pH at equivalence point
e.) pH at half the volume required to reach equivalnce
f.) pKa of acetic acid
Thank you!
Explanation / Answer
a) Molarity of NaOH= 21.51ml, moles of NaOH at equivalence point = molarity* Volume (L) = 9.878*10-2*21.51/1000=0.002125 moles
b) The reaction between CH3COOH and NaOH is CH3COOH + NaOH----> CH3COONa + H2O
moles of acetic acid =0.002125 concentration = 0.002125*1000/20 =0.1062 moles of sodium acetate formed = 0.002125 moles and concentration = 0.002125*1000/21.5+20)=0.0512M
c) CH3COO- + H2O -----------> CH3COOH + OH-
initially 0.0512 0 0
change -x x x
eq. 0.0512-x x x
x2/(0.512-x)= Kb= Kw/Ka, Pka of acetic acid =4.75, Ka=
10-14/ 1.8*10-5 =5.55*10-10 = x2/ (0.0512-x), when solved, x= 5.72*10-6, pOH=5.24, pH= 14-5.24= 8.76
half the volume of NaOH= 21.51/2= 10.755, moles of NaOH= 9.878*10-2*10.755/1000=0.001062
moles of CH3COOH consumed = 0.001062, moles of CH3COOH remaining = 0.002125-0.001062= 0.001063
moles of CH3COONa formed =0.001063
pH= 4.75 + log (0.001062/0.001063)= 4.56