Consider a weak acid-strong base titration in which 25 mL of 0.120 M of acetic a
ID: 947741 • Letter: C
Question
Consider a weak acid-strong base titration in which 25 mL of 0.120 M of acetic acid is titrated with 0.120 M of NaOH.
a) Calculate the pH of the acetic acid solution BEFORE addition of NaOH (pKa of acetic acid = 4.75).
b) Calculate the pH after the addition of 3.00 mL of NaOH.
c) Calculate the pH after the additon of 12.5 mL of NaOH. Notice that this is the half neutralizatiom point: some of the acetic acid molecules are converted to acetate ions producing a buffer whose pH depends on the base/acid ratio (CH3COO-/CH3COOH).
d) Calculate the pH after the addtion of 25 mL of NaOH (equivalence point).
e) Calculate the pH after the addition of 35 mL of NaOH.
f) suggest an indicator other then phenolphthalein that would be suitable for this titration and explain why.
Thank you very much.
Explanation / Answer
Titration
a) before addition of base
CH3COOH <==> CH3COO- + H+
let x amount has dissociated
pKa = -logKa
Ka = 1.80 x 10^-5
Ka = [CH3COO-][H+]/[CH3COOH]
1.80 x 10^-5 = x^2/0.120
x = [H+] = 1.47 x 10^-3 M
pH = -log[H+] = 2.83
b) after 3 ml of NaOH added
CH3COOH + NaOH ---> CH3COONa + H2O
remaining [CH3COOH] = (0.120 M x 22 ml)/28 ml = 0.0943 M
formed [CH3COONa] = (0.120 M x 3 ml)/28 ml = 0.013 M
pH = pKa + log([CH3COONa]/[CH3COOH])
= 4.75 + log(0.013/0.0943)
= 3.89
c) after 12.5 ml of NaOH added
This is half equialence point
[CH3COOH] = [CH3COONa]
pH = pKa = 4.75
d) after 25 ml of NaOH added
This is equivalence point
formed [Ch3COONa] = 0.120 M x 25 ml/50 ml = 0.06 M
salt hydrolyses
CH3COO- + H2O <==> CH3COOH + OH-
let x amount has hydrolyzed
Kb = Kw/Ka = 1 x 10^-14/1.8 x 10^-5 = [CH3COOH][OH-]/[CH3COO-]
5.55 x 10^-10 = x^2/0.06
x = [OH-] = 5.77 x 10^-6 M
pOh = -log[OH-] = 5.24
pH = 14 - pOH = 8.76
e) after 35 ml of NaOH added
excess [NaOH] = 0.120 M x 10 ml/60 ml = 0.02 M
pOH = -log(0.02) = 1.70
pH = 14 - pOH = 12.30
f) The best indicator would be one whose pH lies near equivalence point of titration.
Here equivalence poijt is 8.76, which match with thymol blue (pH range 8.0-9.6), it is yellow in acidic and blue in basic medium. So this indicator can work here.