Question

A 100 mL solution of 0.100 M hydroxyacetic acid (Ka = 1.48 x 10-4 ) is titrated with 0.0500 M KOH. Answer the questions that follow;

(a) What volume of KOH is required to reach the equivalence point?

(b) What is the pH at the equivalence point?

(c) Which indicator below would be suitable for determining the end point? Methyl yellow (pKa 3.1), chlorophenol red (pKa 6.2), thymol blue (pKa 8.9)

Answer 1

A 100 mL solution of 0.100 M hydroxyacetic acid (Ka = 1.48 x 10-4 ) is titrated with 0.0500 M KOH. Answer the questions that follow;

(a) What volume of KOH is required to reach the equivalence point?

Solution :- Reaction equation

_{C2H4O3 + NaOH ----} > C2H3O3Na + H2O

Mole ratio is 1 : 1

Lets calculate the moles of Acid

Moles =molarity x volume in liter

Moles of acid = 0.100 mol per L * 0.100 L = 0.0100 mol

Since the mole ratio is 1 : 1 therefore moles of NaOH needed to reach the equivalence point are same

Hence moles of NaOH needed = 0.010 mol

Now using the molarity and moles of NaOH we can find the volume of NaOH

Volume = moles / molarity

             =0.01 mol /0.0500 mol per L

             = 0.200 L

0.200 L * 1000 ml / 1 L = 200 ml

Therefore it need 200 ml NaOH to reach the equivalence point

(b) What is the pH at the equivalence point?

Solution :- As we calculated in the part a the volume of NaOH needed to reach the equivalence point is 200 ml

Therefore the total volume at the equivalence point is 100 ml + 200 ml = 300 ml

At the equivalence point all the acid is converted to conjugated base

Therefore molarity of the conjugate base = moles / volume

                                                                  = 0.01 mol / 0.300 L

                                                                  = 0.0333 M

Conjugate acts as weak base and produces the OH^- ions

Lets calculate the equilibrium concentration of the OH^-

C2H3O3^- + H2O ---- > C2H4O3 + OH^-

0.0333 M                               0                 0

-x                                          +x                +x

0.0333-x                                x                   x

Using the given ka we can find the kb

Kb = kw/ ka

      =1*10^-14 / 1.48*10^-4

     = 6.76*10^-11

Kb=[C2H4O3][OH-]/[C2H3O3^-]

6.76*10^-11 = [x][x]/[0.0333-x]

We can neglect the x from denominator since Kb is very small

6.76*10^-11 = [x][x]/[0.0333]

6.76*10^-11 * 0.0333 = x^2

2.25*10^-12 = x^2

Taking square root of both sides we get

1.50*10^-6 = [x]= [OH^-]

Now using the OH- concentration we can find the pOH

pOH= -log [OH-]

pOH = - log [1.50*10^-6]

pOH= 5.82

pH + pOH= 14

pH= 14- pOH

pH= 14 – 5.82

pH= 8.18

Therefore at the equivalence point pH = 8.18

(c) Which indicator below would be suitable for determining the end point? Methyl yellow (pKa 3.1), chlorophenol red (pKa 6.2), thymol blue (pKa 8.9)

Solution :-

The pH at the equivalence point is 8.18 which is determined by the indicator thymol blue because it have pka 8.9

Therefore thymol blue can be used as the indicator.