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Example Write a charge balance equation for a solution containing KI and AlI 3 . Solution KI g K + + I - AlI 3 = Al 3+ + 3 I - H 2 O D H + + OH -PowerPoint Presentation

Example Write a charge balance equation for a solution containing KI and AlI 3 . Solution KI g K + + I - AlI 3 = Al 3+ + 3 I - H 2 O D H + + OH -

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Example Write a charge balance equation for a solution containing KI and AlI 3 . Solution KI g K + + I - AlI 3 = Al 3+ + 3 I - H 2 O D H + + OH -

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Write a charge balance equation for a solution containing KI and AlI3.

Solution

KI g K+ + I-

AlI3 = Al3+ + 3 I-

H2O D H+ + OH-

The equation can be written directly considering that the concentration of the charge on Al3+ is three times aluminum concentration, therefore we have:

[I-] + [OH-] = [H+] + [K+] + 3[Al3+]

Equilibrium Calculations Using the Systematic Approach

This section will only be briefly presented since it will not be our approach in solving equilibrium problems. T depends on building up relations as the equilibrium constant, solubility product, autoprotolysis, mass balance, charge balance, etc.. so as to produce equations at least equaling the number of unknowns. The systematic approach is not easily presented in a part of a lecture but needs enough time for mastering in order to be able to solve equilibrium problems effectively. Therefore, we are justified to keep it as simple as possible in our discussion.

Use the systematic approach to calculate the equilibrium concentrations of A, B, and AB in a 0.10 M AB solution. The equilibrium constant is 3.0x10-6.

AB D A + B

Solution

We do not have charges in the chemical equation; therefore do not attempt to write a charge balance equation.

K = [A][B]/[AB]

The mass balance equation can be formulated as:

CAB = [AB] + [A]

0.10 = [AB] + [A]

From the value of the equilibrium constant we know that [AB] >> [A]

Therefore, [AB] = 0.10

Also, AB dissociates into A and B only, therefore [A] = [B]

Substitution in the equilibrium constant expression gives

3.0x10-6 = [A]2/0.10

[A] = 5.5x10-4 M

The relative error = (5.5x10-4 / 0.10) x 100 = 0.55% which means that A is really much smaller than AB

[A] = [B] = 5.5x10-4 M

[AB] = 0.10 -5.5x10-4~ 0.10 M

Using the systematic approach Calculate the equilibrium concentrations of A, B, and AB in a 0.10 M AB (equilibrium constant is 3.0x10-6) in presence of 0.20 M MB (strong electrolyte)where the equilibrium below takes place.

AB D A+ + B-

MB g M+ + B-

CB- = [B-] from strong electrolyte + [B-]from dissociation of AB

[B-]from dissociation of AB = [A+]

CB- = [B-] from strong electrolyte + [A+]

A+ concentration is negligible as compared to amount of B- from strong electrolyte due to very small K and common ion effect seen earlier. Therefore we have

[B-] = 0.20 M

We also have from mass balance that

CAB = [AB] + [A+]

The same argument can also be used where the very low K suggests that [AB] >> [A+], therefore:

0.10 = [AB]

Now, substitution in the equilibrium constant expression ew get:

K = [A+][B-]/[AB]

3.0x10-6 = [A+] * 0.20/0.10

[A+] = 1.5x10-6 M

The relative error = (1.5x10 suggests that -6/0.10) x 100 =1.5x10-3 %

Therefore, the concentrations of the different species are:

[A+] = 1.5x10-6 M

[B] = 0.20 + 1.5x10-6~ 0.20 M

[AB] = 0.10 – 1.5x10-6~ 0.10

Activity and Activity Coefficients suggests that

In presence of strong electrolytes of diverse ions, weak electrolytes tend to dissociate more. This behavior is attributed to the shielding effects exerted by the ions of the strong electrolyte on the ionic species of opposite signs from the weak electrolyte. The result is extra dissociation of the weak electrolyte. The concentration of an ion, from a weak electrolyte, in presence of strong electrolyte is called the activity of the ion and is defined by the relation:

ai = Ci fi

Where ai is the activity of the ion i, Ci is the concentration of ion i, and fi is the activity coefficient of ion i.

Properties of the activity Coefficient suggests that

- The activity coefficient of uncharged species is unity.
- The activity coefficient of ions in dilute solutions (less than 10-4 M) is very close to unity and in fact can be approximated to unity.
- As the concentration of electrolyte (diverse ions) increases, the activity coefficient decreases and activity becomes less than concentration
- The activity coefficient is highly dependent on the charge of diverse ions where as the charge increases, the activity coefficient decreases significantly

Ionic Strength suggests that

A property which represents the strength of ions in solution can be seen using the term ionic strength (m) which can be defined as:

m = 1/2 SCiZi2

Zi is the charge on the ion i. If we look carefully to properties of activity coefficients listed above (3 and 4) it is clear that the term ionic strength combines the effects of concentration and charge of ions and thus will be advantageously used as an important factor affecting fi.

Example suggests that

Calculate the ionic strength of a 0.2 M KNO3 solution.

Solution

m = 1/2 SCiZi2

m = 1/2 (CK+ Z2 K+ + CNO3- Z2NO3-)

m = 1/2 (0.2 x 12 + 0.2 x 12) = 0.2

Example suggests that

Calculate the ionic strength of a 0.2 M K2SO4 solution.

Solution

m = 1/2 SCiZi2

m = 1/2 (CK+Z2 K+ + CSO42- Z2SO42-)

m = 1/2 (2 x 0.2 x 12 + 0.2 x 22) = 0.6

Compare between the value of the ionic strength in this and previous example and see the effect of charge on the answer.

Calculate the ionic strength of the solution containing 0.3 M KCl and 0.2 M K2SO4.

m = 1/2 SCiZi2

m = 1/2 (CK+Z2 K++ CCl- Z2Cl- + CSO42- Z2SO42-)

m = 1/2 (0.7 x 12 + 0.3 x 12 +0.2 x 22) = 0.9

We have substituted 0.7 for CK+ since this is the overall potassium ion concentration but we could make it the sum of two terms:

{ CK+Z2 K+}from KCl + { CK+Z2 K+}from K2SO4which will also give the same result.

Calculation of Activity coefficients M KCl and 0.2 M K

The activity coefficients of an ion i can be calculated provided that three pieces of information are known. These are:

- The ionic strength of the solution
- The charge of the ion
- The effective diameter of the hydrated ion (ai) in angstrom (Ao)
The Debye-Huckel equation can be used for such a calculation where:

- log fi = (0.51 Zi2m1/2)/(1 + 0.33 aim1/2)

In situations where the effective diameter of the ion is 3 A M KCl and 0.2 M Ko (you can assume this situation to solve assigned problems) we can write

- log fi = (0.51 Zi2m1/2)/(1 + m1/2)

However the Debye-Huckel equation gives an approximate value for the activity coefficient. A better relation is the Davies modification of the Debye-Huckel equation. The equation is:

- log fi = (0.51 Zi2m1/2)/(1 + 0.33 aim1/2) – 0.10 Zi2m

In our calculations we will only use the Debye-Huckel equation for estimation of activity coefficients.

Example M KCl and 0.2 M K

Calculate the activity coefficients of K+ and SO42- in a 0.002 M K2SO4 assuming an effective diameter of 3 Ao for both ions.

Solution

To apply Debye-Huckel equation we need to determine the ionic strength of the solution

m = 1/2 SCiZi2

m = 1/2 (0.002 x 2 x 12 + 0.002 x 22) = 0.006

- log f M KCl and 0.2 M Ki = (0.51 Zi2m1/2)/(1 + m1/2)

- log fK+ = (0.51 x 12 0.0061/2)/(1 + 0.0061/2) = 0.037

fK+ = 10-0.037 = 0.918

For sulfate we have

- log fSO42- = (0.51 x 22 x 0.0061/2)/(1 + 0.0061/2) = 0.147

fSO42- = 10-0.147 = 0.71

If we repeat the calculation for a more concentrated solution (0.02 M K2SO4) the ionic strength will be 0.06 and the activity coefficients for K+ and SO42- will be 0.79 and 0.38, respectively. Therefore, you should appreciate the effect of concentration on both ionic strength and activity coefficients as well.