Question

Trichloroacetic acid dissociates partially in water according to the equilibrium below. What is the percentage of molecules dissociated in a 1.00 molal solution of trichloroacetic acid in water if the measured freezing point is -2.53C? The freezing point and Kf for water are 0.00C and 1.86 C-kg/mol, respectively.

CCl3CO2H(aq) -> <- (Equalibrium arrows) CCl3CO2 1- (aq) + H+ (aq)

How do I set up and Solve

Answer 1

Ans. Using dT_f = i K_f m                     - equation 1

            where, i = Van’t Hoff factor = ?

                        K_f = molal freezing point depression constant of the solvent

                        m = molality of the solution

                        dT_f = Freezing point of pure solvent – Freezing point of solution

                                    = freezing point depression

Given,

            m = 1

            Freezing point of solution = - 2.53⁰C

            Freezing point of solvent (water) = 0⁰C

            Kf = 1.86⁰Ckgmol^-1

Now, dTf = Freezing point of pure solvent – Freezing point of solution

            = 0⁰C – (- 2.53⁰C) = 2.53⁰C

Putting the values in equation 1-

2.53⁰C = i x (1.86⁰Ckgmol^-1) x 1.0 m

Or, 2.53⁰C = i x (1.86⁰C m^-1) x 1.0 m                      ; [1 m = 1 mol/ kg; so, kg/mol = m^-1]

Or, 2.53 = 1.86 i

Or, I = 2.53 / 1.86 = 1.36

Thus, Van’t hoff factor for trichloroacetic acid = 1.36

That is, 1 molecule of trichloroacetoic acid shall dissociates into 1.36 dissociated chemical species.

Note that dissociation of a molecule is always in a whole number value, i.e, a molecule can dissociate into 2, 3, 4 or other whole number species but not in fraction like 0.2 or 0.54, etc.

## So, consider the van’t hoff facor = 1.36 as follows-

Let the total number of molecules of trichloroacetic acid (TCAA) = 100

So, 100 (TCAA) dissociated to produce a total of 136 chemical species.

Given,

1 (TCAA) dissociates into 2 ions – TCAA^- and H⁺.

Let the number of (TCAA) dissociated = Y

So, remaining number of un-dissociated (TCAA) = (100 – Y)

Total number (dissociated + Un-dissociated) chemical spices = (2 x Y) + (100- Y)

            Where (2 x Y) indicates that 1 mol TCAA dissociates into 2 ions.

Or, (2 x Y) + (100- Y) = 136

Or, 2Y + 100 – Y= 136

Or, Y = 136 – 100 = 36

That is, a total of 36 TCAA dissociated into (TCAA^- + H⁺) per total 100 molecules TCAA initially present in the solution.

## Again, Coming to i = 1.36

Moles of TCAA dissociated = (1.36 – 1.0) = 0.36

% dissociation = (moles of TCAA dissociated / Initial moles of TCAA) x 100

                        = (0.36/ 1) x 100

                        = 36 %