Equilibrium Lab #2 - Le Châtelier's Principle-The Common Ion Effect

Introduction
The qualitative effects which will be produced in an equilibrium by changing the concentration of one or more of the interacting substances can readily be predicted, based on a kinetic consideration of the way such changes will affect the relative forward and reverse rates. The rate of formation of hydrogen iodide, according to H2 + I2 <---> 2 HI, must be proportional to the number of effective collisions of hydrogen molecules and iodine atoms. If more hydrogen is added to the container, thereby increasing the concentration of hydrogen molecules, there will be more collisions per second, which will thereby increase the rate of the forward reaction. This will momentarily "unbalance" the equilibrium, until the concentration of hydrogen iodide has increased to a point where the opposing reactions again are taking place at equal rates. There will then be more hydrogen, less iodine, and more hydrogen iodide than in the original mixture. The equilibrium has been "shifted to the right".

Conversely, if some iodine is removed from the container, there will be fewer collisions per second between hydrogen molecules and iodine atoms, resulting in a slower forward reaction. The reverse reaction will predominate until a new equilibrium is established with less hydrogen iodide, more hydrogen, and some what less iodine than in the original mixture. Any such concentration changes will affect the "balance" of opposing reactions in accordance with Le Châtelier's Principle. When some stress is applied to a system originally in equilibrium, the system (reaction) automatically will shift in such a direction as to relieve the stress (in this example a change in concentrations) and restore the original conditions as much as possible.

The common ion effect is a special case of the application of the law of chemical equilibrium to ionization reactions. For example, in a solution of the weak base, ammonium hydroxide, there is the equilibrium

NH4OH <-----> NH4+1  +  OH-1

The addition of NH4Cl, (NH4)2SO4, or any other soluble ammonium salt, will increase the concentration of NH4+1, and therefore increase the number of collisions per second between NH4+1 and OH-1. The equilibrium will be shifted to the left, and the concentrations of the OH-1  will be decreased. The NH4+1, since it is common to both the ammonium hydroxide and the added ammonium salt, is called a "common ion".

In the same way, salts that are only slightly soluble can be made even less soluble by increasing the concentration of a common ion. For example, the equilibrium between the slightly soluble salt gypsum, CaSO4.2H2O, and its ions in solution is represented by the equation

CaSO4.2H2O(s) <----->  Ca2+ + SO42- + 2 H2O

The addition of either Ca2+ or SO42-, from any soluble salt containing one of these ions, would shift the equilibrium to the left and decrease the solubility.
 
 

Experimental Procedure
Required Reagents: 0.1 M HC2H3O2, 1 M NH4Cl, 0.1 M NH4OH, 0.1 M BaCl2, 0.20 M Fe(NO3)3,
0.1 M KSCN, 0.0020 M KSCN, 1 M K2CrO4, 1 M NaC2H3O2, 6.0 M NaOH ,2.0 M H2SO4, 5.4 M NaCl(sat.), methyl orange & phenolphthalein indicators.

1. The Shifting of an Equilibrium. The Common Ion Effect.

a) The Chromate Ion-Dichromate Ion Equilibrium
Yellow chromate ion reacts with hydrogen ion to form first hydrogen chromate ion, then by condensation (loss of H2O) orange dichromate ion:

2 CrO42- + 2 H+1  <-----> 2 HCrO4-1    <----->  Cr2O72- + H2O

At present, we need consider only the over-all reaction,

2 CrO42- + 2 H+1  <----->  Cr2O72- + H2O

To 3 mL of 1 M K2CrO4 add several drops of 2 M H2SO4. Mix this, and observe any change. Now add several drops of 6 M NaOH with mixing, until a change occurs. Again add H2SO4. Interpret the observed changes in terms of "shifting the equilibrium."

b) Weak Acids and Weak Bases
Let us examine the relative concentrations of molecules and ions in weak acids and weak bases from the equilibrium point of view.

To 3 mL of 0.1 M HC2H3O2 add a drop of methyl orange, then add 1 M NaC2H3O2, a few drops at a time, with mixing. Explain your observations in terms of the equilibrium equation and the "common ion" effect.

To each of two 3 mL samples of 0.1 M NH4OH add a drop of phenolphthalein. To one sample add 1 M NH4Cl, a few drops at a time, with mixing. To the other add 6 M HCl, a drop at a time, with mixing. In each case, note any changes in colour and in odour of the solutions. Write the equation for the equilibrium in dilute NH4OH and interpret the different observed results, including the additional overall net ionic equation for the reaction with HCl, in terms of shifting the equilibrium.

c) Saturated Solution Equilibria
(1) NaCl(s). The solubility of sodium chloride at 20oC is 358 g/1000 g H2O, or 358g/1358 g of solution, which has a density of 1.200 g/mL. Verify this as 5.4 M NaCl. Write an equation for the ionic equilibria present in a saturated NaCl solution. Predict what would happen when 10 mL of saturated NaCl is treated with a few millilitres of concentrated 12 M HCl. Try it.

(2) BaCrO4(s). To 3 mL of 0.1 M BaCl2, add a few drops of 1 M K2CrO4, and then a little 6 M HCl. Explain your observations in terms of the equilibrium equations involved.
 
 

d) The Thiocyano-iron(III) Complex Ion
Iron(III) ion, Fe3+, and thiocyanate ion, SCN-, react to form a blood-red substance in solution.

  Fe3+ + SCN <---->  Fe(SCN)2+

Add 1 mL of 0.2 M Fe(NO3)3 to 2 mL of 0.1 M KSCN, and dilute this until a moderately red colour is attained; about 25 to 35 mL of distilled water will be required. To a 5 mL portion of this, add a little 0.2 M Fe(NO3)3; to a second 5 mL portion add a little 0.1 M KSCN; and to a third 5 mL portion add a few drops of 6 M NaOH. (Fe(OH)3 is quite insoluble.) Correlate your observations with Le Châtelier's principle and the above equilibria. Do your data prove the correctness of the above equation?
 

2. A Quantitative Study of the Equilibrium Between Iron(III) Ion and Thiocyanate Ion
We shall determine colourmetrically the concentrations of the constituents in the Equilibrium involving Fe3+ and SCN- which you studied qualitatively in 1(d) above. Form the data obtained we can calculate an equilibrium constant according to the equation.

You will first prepare a standard comparison solution of red thiocyano-iron(III) complex of known concentration by mixing a known low concentration of SCN- ion with a large excess of Fe3+ ion. You can assume essentially complete reaction, so that the concentration of the red complex can be calculated as equal to the total SCN- concentration in the mixture.

You will then prepare a number of test tube mixtures, with successively lower Fe3+ concentration, resulting in a corresponding shift of the equilibrium and a corresponding decreased concentration, and colour, of the red complex. We can measure this decreased concentration by removing solution from the first known standard sample until the colours, viewed lengthwise through the tubes, exactly match. The concentrations will then be inversely proportional to the respective depths of solution.

Solution Preparation
Select five clean 15-cm test tubes of the same diameter, and label them 1 to 5. Into each, carefully measure 10.0 mL of 0.0020 M KSCN, using a 10 mL pipet or a 10 mL graduated cylinder. (Rinse this first with the solution. A disposable pipet is useful in adjusting the final volume.) To test tube No. 1, add 10.0 mL of 0.20 Fe(NO3)3. Now dilute 5.0 mL of 0.20 M Fe(NO3)3 to a total of 50.0 mL, mix this well, and measure precisely 10.0 mL into test tube No 2. Save precisely 25.0 mL of the remaining diluted Fe3+ solution, and dilute this further to 50 mL total volume. Mix this well, and add 10.0 mL of it to test tube No. 3. Continue to make similar 2-fold dilutions for the remaining Fe3+ solution, and add the successive 10.0 mL portions respectively to test tubes No. 4 and 5. Mix the contents of each test tube by very gentle swirling.

Colourmetric Concentration Measurements
Determine the concentration of the red-cyano-iron(III) complex in each test tube No. 2-5, by comparison with the standard known concentration in test tube No. 1, as follows. First prepare two cylindrical paper sleeves, which can be slipped on the test tubes to exclude side lighting, by wrapping strips of paper around a test tube and securing the paper edges with tape. Arrange test tubes No. 1 and 2 so as to look lengthwise through the solution toward a good diffuse light source.

Remove small amounts of standard solution from test tube No. 1 (putting this into a small clean beaker or Erlenmeyer flask, since you may need it again), until the colour intensities appear exactly the same when viewed lengthwise through the tubes. Reverse the viewing positions of the tubes for better comparison and make final adjustments of the solution in test tube No. 1 with a dropper pipette. Measure the depth of each solution to the nearest millimetre. Repeat this comparison procedure with test tubes No. 1 and 3, No. 1 and 4, and No.1 and 5.

Calculate your quantitative observations so as to evaluate the possible equilibrium involved, in the following way. First write the equilibrium constant expression, assuming that the red complex is Fe(SCN)2+, from the equation given. Calculate the initial concentrations (before reaction) of Fe3+ and of SCN- in each of the test tubes. (Express concentrations to two significant figures, in exponential notation, such as 5.0 x 10-4) Assume that in tube No. 1 all the SCN- reacts to form Fe(SCN)2+; then, from the relative depth measurements in each tube No.2-5, compared with tube No. 1, calculate the equilibrium concentration of Fe(SCN)2+ in each tube. You can then calculate the equilibrium concentrations of Fe3+ and of SCN-, by subtracting this Fe(SCN)2+ concentration from the respective Fe3+ and SCN- initial concentrations for each tube No. 2-5. Finally calculate the value of Keq for each mixture No. 2-5.
 
 
 
 
 
 
 
 

Lab Report    Reversible Reactions and Chemical Equilibrium

1. The Shifting of an Equilibrium, The Common Ion Effect

a) The Chromate Ion-Dichromate Ion EquilibriumRewrite the equation for this equilibrium
 
 

Describe the colour changes obtained on the addition the H2SO4 and NaOH, respectively, and interpret these effects.
 
 
 
 
 
 

b) Weak Acids and Weak Bases

The equation for the equilibrium in dilute acetic acid is:
 
 

The equation for the equilibrium in dilute ammonium hydroxide solution is:
 
 

Note any observed odour or colour changes when 0.1 M NH4OH (+ phenolphthalein) is treated with:

NH4Cl ________________________ HCl _______________________

Which direction, right or left, does each reagent shift the above equilibrium? Explain fully.
 
 
 
 
 
 

c) Saturated Solution Equilibria

(1) NaCl. Calculate the molarity of saturated NaCl, bases on the data in the experimental directions.

______ M NaCl

Write an equation for the equilibrium present in saturated NaCl, and explain your observations when concentrated HCl was added to this equilibrium.
 
 

(2) BaCrO4 Write the equation for the equilibrium established when a few drops of 1 M K2CrO4 is added to 3 mL of 0.1 M BaCl2, and explain any changes observed when this is treated with a little 6 M HCl.
 
 
 
 
 

(d) The Thiocyano-iron(III) Complex Ion

The equilibrium equation is Fe3+ + SCN-1   <------>  Fe(SCN)2+

Explain fully, in terms of Le Châtelier's principle, the effect on those equilibria of adding (1) 0.2 M Fe(NO3)3, (2) 0.1 M KSCN, (3) 6 M NaOH.
 
 

2. A Quantitative Study of the Equilibrium Between Iron(III) Ion and Thiocyanate Ion

For the equilibrium   Fe3+ + SCN-1  <------>  Fe(SCN)2+

Solution         Depth                              Initial [ ]'s                       Equilibrium [ ]'s
 

Test Tube No.  Tube  No. 1 Other Tube Fe3+ SCN- Fe(SCN)2+ Fe3+ SCN-  Keq