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Chapter 15 Acids and Bases Reactions

Chapter 15 Acids and Bases Reactions. Objectives. 15.1 Distinguish the overall, ionic, and net ionic equations for an acid base reaction 15.1 Classify acids and bases using the hydrogen transfer definition 15.1 Predict and explain the final results of an acid base reaction. Objectives.

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Chapter 15 Acids and Bases Reactions

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  1. Chapter 15 Acids and Bases Reactions

  2. Objectives • 15.1Distinguish the overall, ionic, and net ionic equations for an acid base reaction • 15.1 Classify acids and bases using the hydrogen transfer definition • 15.1 Predict and explain the final results of an acid base reaction

  3. Objectives • 15.2 Evaluate the importance of a buffer in controlling pH • 15.2 Design strategies for doing acid base titrations, and calculate results from titration data

  4. Acid and Base Reactions • Are called Neutralization Reactions. • In most cases, water and a salt is formed • NaOH + HCl Water and NaCl • Net Ionic: Na+ + OH- + H+ + Cl-  Water • Spectators: Na+ and Cl-

  5. Equivalence Point • At the equivalence point, equal amounts of acid and base have been added together, so there are no reactants left. • In the case of adding HCl (a strong acid) to NaOH (a strong base), at the equivalence point, there is only water and a neutral salt. • HCl + NaOHNaCl + H2O

  6. Titration of SA with SB • In the titration of a strong acid with a strong base, at the equivalence point, there are only salts and water in the solution. • The salts produced are neutral (the conjugate of a strong acid or base is neutral), so the pH of the solution is 7.0 (also 7.0 on the pOH scale)

  7. Graphical Representation of Titrations • For a strong base being added to a strong acid, the solution (acid) starts off very low on the pH scale. • It stays low until it gets very close to the equivalence point. This is where it jumps up. Once all of the acid has been eliminated, the base is present but not the acid, so the pH is very high.

  8. Strong Base added to Strong Acid

  9. What if a Strong Base was added to a Weak Acid solution? • Beginning will be much closer to a pH of 7 • At the equivalence point, we again have salts and water. • However, the conjugate for the Weak acid is NOT neutral, it is actually basic, so the solution will be basicat the equivalence point for a strong base-weak acid titration

  10. Strong Base added to Weak Acid

  11. Conjugate Strength • The conjugate base of a weak acid is a weak base. • At the equivalence point, the weak acid is all gone, but the conjugate base is present. • The pH of the solution will be greater than 7 due to the weak base. • And same goes for the conjugate of a weak base, it will be acidic at the equivalence point

  12. Verification of the pH • 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution. • a) 0 mL of acid has been added • b) 20 mL of acid has been added • c) 39 mL of acid has been added • d) 40 mL of acid has been added • e) 41 mL of acid has been added • f) 60 mL of acid has been added

  13. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution. • a) 0 mL of acid has been added • The solution is 0.1 M NaOH. • - Log (OH-) = pOH • pOH = 1 • 14 – 1 = pH • pH = 13

  14. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution • b) 20 mL of acid has been added • How many mols react? • Molarity x Volume = Mols • 0.1 M (HCl) x 0.020 L = 0.002 mols of HCl • Were added to the solution • Original Base was • 0.1 M (NaOH) x 0.040 L = 0.004 molsNaOH

  15. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution • Subtract the Acid and Base from one another • 0.004 NaOH – 0.002 HCl = 0.002 NaOHmols • Calculate new molarity • Find new volume 40 mL + 20 mL = 0.060 L • Find new molarity 0.002 NaOH / 0.060 = 0.033 • - log (0.033) = 1.477 pOH • 14 – 1.477 = 12.522

  16. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution • c) 39 mL of acid has been added • Original Base = 0.004 mols base • Acid added = Molarity x Volume • 0.1 x 0.039 = 0.0039 mols acid • Subtract acid from base • 0.004 – 0.0039 = 0.0001 mols Base • New volume = Add volumes • 0.04 + 0.039 = 0.079 • Find Molarity (Mols / Volume) • 0.0001 mols / 0.079 = 0.00126 M • Determine pOH = - Log (0.00126) = 2.9 (pH of 11.1)

  17. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution • d) 40 mL of acid has been added • Original Base = 0.004 mols base • Acid added = Molarity x Volume • 0.1 x 0.04 = 0.004 mols acid • Subtract acid from base • 0.004 – 0.004 = 0.0000 mols • NO ACID OR BASE = pH 7

  18. 40 mL of a 0.1 M NaOH solution is titrated with 0.1 M HCl solution • e) 41 mL of acid has been added • Original Base = 0.004 mols base • Acid added = Molarity x Volume • 0.1 x 0.041 = 0.0041 mols acid • Subtract base from acid (More acid) • 0.0041 (acid) – 0.004 = 0.0001 molsAcid • New volume = Add volumes • 0.04 + 0.041 = 0.081 • Find Molarity (Mols / Volume) • 0.0001 mols / 0.081 = 0.00123 M • Determine pH = - Log (0.00123) = 2.9

  19. Buffers • Resist pH changes • Composed of a solution with Weak acids and Weak Bases • Strong Acids and Strong Bases do NOT resist pH changes

  20. How they resist • Consider a solution with • Acetic acid (HC2H3O2) • Acetate (C2H3O2-) Which is the conjugate • If a Base is added, it reacts with the Acetic acid • If an Acid is added, it reacts with the Acetate

  21. Your Blood • Has an acid/conjugate base buffer to protect you • pH of blood is 7.4. A change in even a few tenths means serious illness or death

  22. Your Blood Buffer • The following equilibrium serves as a buffer for your blood • H2CO3 (aq) + H2O (l)  • HCO3- (aq) + H3O+ (aq) • Carbonic acid (H2CO3) and its CB form the basis of the buffer in our blood.

  23. Two Equilibriums Coexisting • Breathing rate dictates the amount ofcarbon dioxide in the blood • The carbonic acid is also in equilibrium with • H2CO3 (aq) H2O (l) + CO2 (g) • Both of these equilibriums maintain our blood.

  24. Your Blood Buffer • When our blood pH gets too high, we breathe out more carbon dioxide. This affects the second equilibrium • Breathe Out, Decrease • H2CO3 (aq)  H2O (l) + CO2 (g)

  25. Your Blood Buffer • Since we remove products, we shift to form more, this causes the carbonic acid to decrease • Since the amount of carbonic acid decrease, the other reaction shifts left

  26. Your Blood Buffer • Decrease • H2CO3 (aq) + H2O (l) •  HCO3- (aq) + H3O+ (aq) • And the pH rises as the concentration of hydronium becomes smaller • Factoid: After a marathon, many runners have lower pH’s then 7.4. This is because of high carbon dioxide concentrations in the blood. • Why do people hyperventilating breathe in a brown bag?

  27. pH Calculation of Buffer • Acetic Acid has a Ka of 1.8 x 10-5 • If a solution is 1.0 M Acetic acid and 1.0 M Acetate (CB), what is its pH? • Ka = x2 / M in the past, but now one of the x’s on top is the CB • Ka = 1x / 1

  28. pH Calculation of Buffer • What if that last solution had some base added to it? Which concentration goes down? Which concentration goes up? • What is the new pH if the one that went down is now 0.5 M?

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