CHEMICAL EQUILLIBRIUM LAB Experimental Procedure System 1: Cro. on/Cr20, 2 lon E
ID: 480585 • Letter: C
Question
CHEMICAL EQUILLIBRIUM LAB
Experimental Procedure System 1: Cro. on/Cr20, 2 lon Equilibrium 1. Measure 3 mL of 0.1M K2Cro4 solution. 2. Pour the solution in a clean test tube. 3. Record the color of the solution. 4. While stirring the solution, add 1 drop of 3M H2SO4 to the 0.1M K2Cro4 solution in the test tube. 5. Continue stirring and adding 3M H2So4 dropwise until you observe a color change in the equilibrium mixture. 6. Record your observations. 7. To the same test tube, stir and add 1 drop of 1M NaoH. 8. Continue stirring and adding 1M NaoH dropwise until you observe a color change in the equilibrium mixture. 9. Record your observations. 10. Discard the mixture into the Hazardous Waste Container. System 2: Dissociation of CHaCOOH in H20 11. obtain 2 clean test tubes. a. Test Tube 1 i. Measure and pour 3 ml 1M CH3COOH solution in the test tube. ii. Add 3 drops of methyl orange indicator to the test tube. iii. Stir the solution iv. Record your observations. v. Add 2 or 3 crystals of CHscoONa using a micro spatula to the test tube. vi. Stir the solution until the crystals dissolve. vii. Continue to add individual crystals and stir until you observe a change in the equilibrium mixture. viii. Record your observations. b. Test Tube 2 i. Measure and pour 3 mL 1M CH3COOH solution in the test tube. ii. Add 3 drops of methyl orange indicator to the test tube. iii. Stir the solution. iv. Add 1 drop of 1M NaoH to the test tube v. Stir the solution. vi. Continue stirring and adding 1M NaoH dropwise until you observe a change in the equilibrium mixture. Record your observations. 12. Discard the mixtures in test tubes 1 and 2 into the Hazardous waste container. system 3: Cocl lon/DCocl413 lon Equilibrium 13. Half fill a 250 ml beaker with tap water 14. Turn on the hot plate all the way and heat the water. (You will use this hot waterfor system 4) 15. easure 3 mL of 0.1M CoCl2 and pour the solution in a clean test tube 16. Record the color of the solutionExplanation / Answer
System 1: CrO42-/Cr2O72- Ion Equilibrium
Net ionic equation:
2 CrO42- + 2 H+ <======> Cr2O72- + H2O
Keq = [Cr2O72-][H2O]/[CrO42-]2[H+]2 ……(1)
H2SO4 is a strong electrolyte and dissociates into H+ and SO42- (doesn’t participate in the reaction) as below:
H2SO4 ------> 2 H+ + SO42-
As per the expression for Keq (1) above, addition of H2SO4 increases [H+]. The denominator increases. As temperature remains unchanged, Keq must remain constant. This is possible only when the numerator increases and the only way to increase the numerator is to produce more Cr2O72- so that [Cr2O72-] increases. Therefore, addition of more H2SO4 will shift the reaction to the right.
The shift of the equilibrium reaction toward right is apparent from the color change of yellow CrO42- solution to orange Cr2O72- solution.
The effect of addition of H2SO4 has already been explained above. In short, addition of H2SO4 increases [H+] and thus more Cr2O72- is formed to counter-balance the effect of increased [H+].
Addition of NaOH will shift the equilibrium to the left.
The effect of addition of NaOH is observed from a change of orange Cr2O72- to yellow CrO42-.
Consider the reaction of NaOH (OH- actually) with H+:
H+ + OH- -------> H2O
Addition of NaOH lowers the available H+ by forming water. This reduces the [H+] and increases [H2O] in expression (1) above. As Keq must remain constant, hence, the reaction of CrO42- with H+ is suppressed and the reaction of Cr2O72- with water (to produce CrO42-) is enhanced. This increases the concentration of CrO42- and hence the color changes to yellow.
System 2: Dissociation of CH3COOH in water
Dissociation of CH3COOH in water:
CH3COOH + H2O <=====> CH3COO- + H3O+
Keq = [CH3COO-][H3O+]/[CH3COOH][H2O] ……(2)
The equilibrium will shift to the left on addition of CH3COONa.
A deepening of the color of the solution (to orange) confirms that the equilibrium shifts to the left upon addition of CH3COONa.
CH3COONa dissociates in solution as below:
CH3COONa -------> CH3COO- + Na+
Addition of CH3COONa increases [CH3COO-]. Keq must remain constant as temperature remains constant. This is possible only when [CH3COOH] increases to counteract the increased [CH3COONa]. Therefore, the dissociation of CH3COOH is suppressed and the concentration of CH3COOH increases. Therefore, the solution becomes orange-red in colour.
Addition of NaOH will shift the equilibrium to the left.
The effect of addition of NaOH can be seen from the color of the solution remaining orange.
Consider the reaction of CH3COOH with NaOH as below:
CH3COOH + NaOH -------> CH3COONa + H2O
Addition of NaOH to CH3COOH produces more CH3COONa. As more CH3COONa is produced, [CH3COONa] is increased and hence [CH3COOH] must increase to keep Keq constant. Therefore, the dissociation of CH3COOH is suppressed, thereby increasing the concentration of CH3COOH and hence the color stays orange.