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Applied Le Chatelieir's Principle Lesson 4.

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Presentation on theme: "Applied Le Chatelieir's Principle Lesson 4."— Presentation transcript:

1 Applied Le Chatelieir's Principle Lesson 4

2 Lab 19A Le Chatelier’s Principle Copy the five Equilibrium Equations into your lab handout

3 Part I Thymol Blue (blue) H+ ⇌ Thymol Blue (yellow)

4 Part II Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq) yellow brown
Fe3+(aq) + 3OH- → Fe(OH)3(s)

5 Part III Co(H2O) Cl- ⇌ Co(H2O)4Cl H2O(l) pink blue

6 Part IV Energy + 2CrO42- + 2H+ ⇌ Cr2O72- + H2O(l) yellow orange
Ba2+ + CrO42- → BaCrO4(s)

7 Part V Cu(H2O)42+ + 2NH3 ⇌ Cu(OH)2(s) + 2NH4+ + 2H2O
light blue powder blue Cu(OH)2(s) NH3 ⇌ Cu(NH3) OH- powder blue violet

8 Application of Le Chatelier’s Principle
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ Add CrO42- Add Cr2O72-

9 Application of Le Chatelier’s Principle
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ Add CrO42- Add Cr2O72-

10 Application of Le Chatelier’s Principle
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ Add CrO42- Add Cr2O72-

11 Application of Le Chatelier’s Principle
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ right decrease increase orange Add CrO42- Add Cr2O72-

12 Application of Le Chatelier’s Principle
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ right decrease increase orange Add CrO42- right increase increase ? Add Cr2O72-

13 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Know the following: Any acid is H+ Any base is OH- Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add H+ right decrease increase orange Add CrO42- right increase increase ? Add Cr2O72- left increases increases ?

14 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour 

15 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Acids react with bases H+ + OH- → H2O(l) Stress Shift [CrO42-] [Cr2O72-] colour  Add NaOH

16 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Acids react with bases H+ + OH- → H2O(l) NaOH → Na OH- Stress Shift [CrO42-] [Cr2O72-] colour  Add NaOH

17 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Acids react with bases H+ + OH- → H2O(l) NaOH → Na OH- spectator ion stress ion- lowers H+ Stress Shift [CrO42-] [Cr2O72-] colour  Add NaOH

18 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Acids react with bases H+ + OH- → H2O(l) NaOH → Na OH- spectator ion stress ion- lowers H+ Stress Shift [CrO42-] [Cr2O72-] colour  Add NaOH left increases decreases yellow

19 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange BaCrO4 has low solubility BaCr2O7 has high solubility Stress Shift [CrO42-] [Cr2O72-] colour 

20 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange BaCrO4 has low solubility BaCr2O7 has high solubility Stress Shift [CrO42-] [Cr2O72-] colour  Ba(NO3)2

21 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange BaCrO4 has low solubility- Ba2+ reacts with CrO42- BaCr2O7 has high solubility- no reaction Ba(NO3)2 → Ba NO3- Stress Shift [CrO42-] [Cr2O72-] colour  Ba(NO3)2

22 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange BaCrO4 has low solubility BaCr2O7 has high solubility Ba(NO3)2 → Ba NO3- stress spectator Stress Shift [CrO42-] [Cr2O72-] colour  Ba(NO3)2

23 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange ↓ Ba2+ reacts with this ion BaCrO4 has low solubility BaCr2O7 has high solubility Ba(NO3)2 → Ba NO3- stress spectator Stress Shift [CrO42-] [Cr2O72-] colour  Ba(NO3)2

24 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange ↓ Ba2+ reacts with this ion BaCrO4 has low solubility BaCr2O7 has high solubility Ba(NO3)2 → Ba NO3- stress spectator Stress Shift [CrO42-] [Cr2O72-] colour  Ba(NO3)2 left decreases decreases yellow

25 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange Stress Shift [CrO42-] [Cr2O72-] colour  HCl

26 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange HCl → H+ + Cl- Stress Shift [CrO42-] [Cr2O72-] colour  HCl

27 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange HCl → H+ + Cl- stress spectator Stress Shift [CrO42-] [Cr2O72-] colour  HCl

28 Application of Le Chatelier’s Principle-1
Energy CrO H+ ⇄ Cr2O H2O(l) yellow orange HCl → H+ + Cl- stress spectator Stress Shift [CrO42-] [Cr2O72-] colour  HCl right decreases increases orange

29 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue Stress stress ion spectator ion shift colour  HCl

30 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue HCl → H+ + Cl- Stress stress ion spectator ion shift colour  HCl

31 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue HCl → H+ + Cl- spectator stress Stress stress ion spectator ion shift colour  HCl

32 Application of Le Chatelier’s Principle- 2
Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue HCl → H+ + Cl- spectator stress Stress stress ion spectator ion shift colour  HCl Cl- H+ right blue

33 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue Stress stress ion spectator ion shift colour  Increase temperature

34 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue Stress stress ion spectator ion shift colour  Increase temperature

35 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue Stress stress ion spectator ion shift colour  ↑ temp none none right blue

36 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue Stress stress ion spectator ion shift colour  AgNO3

37 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue AgNO3 → Ag NO3- Stress stress ion spectator ion shift colour  AgNO3

38 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue AgNO3 → Ag NO3- stress- low solubility (data book-4) with Cl- Stress stress ion spectator ion shift colour  AgNO3

39 Application of Le Chatelier’s Principle- 2
Energy + Co(H2O) Cl- ⇄ Co(H2O)4Cl H2O(l) pink blue AgNO3 → Ag NO3- stress- low solubility (data book-4) with Cl- Stress stress ion spectator ion shift colour  AgNO3 Ag+ NO3- left pink

40 Here is a the first question from your lab
Here is a the first question from your lab. I will give you a good answer so that you understand the level of detail that is required in the response 1. Explain the colour changes observed when Fe(NO3)3, KSCN, and NaOH were added to the thiocyanatoiron equilibrium system. Start by writing the equilibrium equation. Fe3+(aq) + SCN-(aq) ⇄ Fe(SCN)2+(aq) yellow brown Adding Fe(NO3)3 increases the [Fe3+] and causes the reaction to shift right and turn brown. Adding KSCN increases the [SCN-] and causes the reaction to shift right and turn brown. Adding NaOH increases the [OH-] which reacts with Fe3+, lowering the [Fe3+] causing the equilibrium to shift left and turn yellow.


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