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Unit 12: Acid and Bases

Unit 12: Acid and Bases. Chapter 19. 4 th Hour. I am gone today so you will go through this Powerpoint to get the notes for objective 38 and the beginning of 39. Remember, there is an open-note quiz tomorrow and I expect a good report from the sub.

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Unit 12: Acid and Bases

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  1. Unit 12: Acid and Bases Chapter 19

  2. 4th Hour • I am gone today so you will go through this Powerpoint to get the notes for objective 38 and the beginning of 39. • Remember, there is an open-note quiz tomorrow and I expect a good report from the sub.

  3. Before we get to the new material, I want to get an idea of how you are doing on the material from the previous few days. • Work on the following problems with the person sitting next to you. • Record your answers on a separate piece of paper (one per group).

  4. Name the following: • HNO3 • Mn(OH)3 • HBr • CuOH • H3PO3

  5. Write the complete dissociation reaction for the following: • HNO3 + H2O  • H2CO3 + H2O 

  6. Calculate the pH for the following. • [H3O-] = 0.0008 M • HCl + H2O  H3O- + Cl- [HCl] = 0.02 M • NaOH Na+ + OH- [NaOH] = 0.036 M

  7. Turn your answers. • At this point, we will begin objective 38. • This objective deals with weak acids and weak bases.

  8. Writing Kaequations • The same principles that apply to equilibrium equations will also apply to acids and bases. • For example: HBr(aq) + H2O(l) H3O+(aq) + Br-(aq) Ka =

  9. Using K equations • Most commonly, these K equations are used to determine the pH of a substance. • This is done by calculating the concentration of the hydronium ion. Ka= • Once that concentration is known, use the pH equation: -log[H3O+]

  10. Example HF(aq) + H2O(l) H3O+(aq) + F-(aq) • Given the following equation, calculate the pH with the following information. • [HF] = 0.5M • Ka = 7.2 x 10-4

  11. Example HF(aq) + H2O(l) H3O+(aq) + F-(aq) • Given the following equation, calculate the pH with the following information. • [HF] = 0.5M • Ka = 7.2 x 10-4 First, write the Ka equation.

  12. Example HF(aq) + H2O(l) H3O+(aq) + F-(aq) • Given the following equation, calculate the pH with the following information. • [HF] = 0.5M • Ka = 7.2 x 10-4 Second, fill in the numbers. Since every HF breaks into H3O+ and F-, then we can assume that they both have the same concentration.

  13. Example HF(aq) + H2O(l) H3O+(aq) + F-(aq) • Given the following equation, calculate the pH with the following information. • [HF] = 0.5M • Ka = 7.2 x 10-4 X2 = 3.6 x 10-4 X = 0.019 M = [H3O+] Third, solve for X.

  14. Example HF(aq) + H2O(l) H3O+(aq) + F-(aq) • Given the following equation, calculate the pH with the following information. • [HF] = 0.5M X = 0.019 M = [H3O+] • Ka = 7.2 x 10-4 -log [0.019] = pH pH = 1.72 Finally, solve for pH.

  15. Kb • It is simple to calculate the pH for acids using a Ka equation because the hydronium ion (H3O+) concentration can be solved for. • For bases though, it is the hydroxide ion that is solved for: Fe(OH)2(aq) Fe+2(aq) + 2OH-(aq) Thus: Kb =

  16. Kb • Once [OH-] is known, it is possible to calculate pOH • pOH= -log[OH-] • Since most bases are aqueous solutions, we can use the water dissociation to complete the problem. • pKw=pH + pOH

  17. Kb Recap • Write the dissociation equation. • Calculate [OH-] from the Kb equation. • Determine pOH • Use the water dissociation to determine pH: • 14 = pH + pOH

  18. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M • [Ca+2] = 0.0136 M • Kb = 3.7 x 10-3

  19. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M • [Ca+2] = 0.0136 M • Kb = 3.7 x 10-3 First, write the Kb equation.

  20. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M • [Ca+2] = 0.0136 M • Kb = 3.7 x 10-3 Second, fill in the numbers.

  21. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M X2 = 1.36 x 10-2 • [Ca+2] = 0.0136 M • Kb = 3.7 x 10-3 x = 0.12 M = [OH] Third, solve for X.

  22. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M x = 0.12 M = [OH] • [Ca+2] = 0.0136 M -log [0.12] = pOH • Kb = 3.7 x 10-3 0.93 = pOH Fourth, solve for pOH.

  23. Example • Ca(OH)2(aq) Ca+2(aq) + 2OH-(aq) • Calculate the pHgiven the following information: • [Ca(OH)2] = 0.05 M 0.93 = pOH • [Ca+2] = 0.0136 M 14 = pH + pOH • Kb = 3.7 x 10-3 14 = pH + 0.93 pH = 13.07 Fifth, solve for pH.

  24. 39 Neutralization • When acids and bases react, they will create a neutralization reaction. • This is because the pH begins to return to 7 (neutral) • Neutralization reactions are essentially double replacement reactions in which the products are always a salt and water. • A salt does not refer to NaCl but rather the product of an acid/base reaction. Acid + Base  Salt + Water

  25. Two Examples • HCl + NaOH H2O + NaCl • 2HBr + Ca(OH)2 2H2O + CaBr2 Acid Base Water Salt Remember to balance the equations

  26. The rest of class is designated to work on your homework packet. • Remember, we have an open-note quiz tomorrow.

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