Question

Consider a titration of 25.00 mL Chloroacetic Acid solution [ka=1.4x10^-3] with 0.1202 M solution of sodium hydroxide. The volume of 27.40 mL of NaOH(aq) was needed to reach the equivalence point. Calculate:

a) The concentration of the chloroacetic acid solution before the titration

b) the pH of the chloroacetic acid solution before 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 1.00mL of the NaOH has been added PAST the equivalence point.

f) the volume (in mL) of 0.1202 M solution of barium hydroxide that would have to be used to reach the equivalence point if it were used instead of the 0.1202M sodium hydroxide solution?

SHOW ALL WORK

Answer 1

Solution :-

Reaction equation

C2H3O2Cl + NaOH --- > C2H2O2ClNa + H2O

Mole ratio of chloroacetic acid with NaOH is 1 : 1

Part a)

Since the mole ratio of the acid and base is 1 :1 then we can use the formula

Concentration of acid = concentration of NaOH * volume of NaOH / volume of acid

                                      = 0.1202 M * 27.40 ml / 25.00 ml

                                      = 0.1317 M

Hence the concentration of the chloroacetic acid is 0.1317 M

Part b)

Using the given Ka we can find the concentration of the H3O^+ as follows

C2H3O2Cl    + H2O    ------ > H3O^+   + C2H2O2Cl^-

0.1317 M                                 0                     0

-x                                             +x                      +x

0.1317 –x                                 x                       x

Ka= [H3O^+][C2H2O2Cl^-] / [C2H3O2Cl]

1.4*10^-3 = [x][x]/[0.1317-x]

1.4*10^-3 *[0.1317-x]=x^2

-0.0014x + 0.000184=x^2

Subtract both sides by x^2 then we get

-0.0014x + 0.000184-x^2=x^2-x^2

-0.0014x +0.000184-x^2 = 0

Writing in the quadratic equation form we get

-x^2 – 0.0014x + 0.000184 =0

Solving for x we get

X= 0.01289 M

Therefore the concentration of the H3O^+ is 0.01289 M

pH= -log [H3O^+]

pH= - log[0.01289]

pH= 1.889

part c) At the half equivalence point half of the acid is converted to conjugate base therefore the concentration of the acid and conjugate base are equal so the ratio of the [Conj Base]/[acid] = 1

At the half equivalence point pH= pka

Pka = -log ka

Pka= -log[1.4*10^-3]

Pka= 2.854

Therefore at the half equivalence point pH= 2.854

Part d) at the equivalence point all the acid is converted to conjugate base

So lets calculate the concentration of the conjugate base at the total volume at the equivalence point

[conj base] = [initial conc. Of base] [initial volume] / [total volume]

                     = [0.1202 M][27.40 ml] / [27.40 ml + 25.00 ml]

                     = 0.06285 M

Conjugate base gives the OH^- ions when it dissociate in water

Lets calculate the concentration of the OH^-

Kb= kw/ka

    = 1*10^-14 / 1.4*10^-3

    = 7.14*10^-12

C2H2O2Cl^-   + H2O   ---- > C2H3O2Cl + OH^-

0.06285 M                                0                    0

-x                                             +x                     +x

0.06285-x                              x                        x

Kb=[C2H3O2Cl][OH-]/[C2H2O2Cl^-]

7.14*10^-12 = [x][x]/[0.06285-x]

Since the kb is very small therefore we can neglect the x from denominator then we get

7.14*10^-12 = [x][x]/[0.06285]

7.14*10^-12 * 0.06285 = x^2

4.49*10^-13 = x^2

Taking square root on both sides we get

6.69*10^-7 = x = [OH-]

pOH=-log[OH^-]

pOH= -log[6.69*10^-7]

pOH= 6.17

pH= 14 – pOH

pH = 14 – 6.17

pH= 7.83

part e ) when the 1.00 ml of NaOH is added after the equivalence point then only OH^- ion remain in the solution because NaOH is strong base and dissociate completely

total volume of solution = 25.00 ml + 27.40 ml + 1.00 ml = 53.4 ml

new molarity of the OH^- = [0.1202 M] * [1.00 ml] / [53.4 ml]

                                                = 0.00225 M

pOH= -log[OH-]

pOH= -log [0.00225]

pOH= 2.65

pH= 14 – pOH

     = 14 – 2.65

    = 11.35

Part f)

Ba(OH)2 is dibasic and NaOH is monobasic

So the mole ratio of the Ba(OH)2 with acid is 2 :1

2 mol acid = 1 mol Base (Ba(OH)2)

Therefore volume of the Ba(OH)2 needed to reach the equivalence point is half than the volume of NaOH

Volume of Ba(OH)2 = volume of NaOH / 2

                                   = 27.40 ml / 2

                                  =13.70 ml