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CHEM 1011

CHEM 1011. pH and Buffer Solutions. pH and Buffer Solutions. Br ø nsted-Lowry Theory Acid-proton donor Base-proton acceptor. pH and Buffer Solutions. For example, HCl + H 2 O ↔ H 3 O + + Cl - For the forward reaction, HCl – proton donor - acid H 2 O – proton acceptor - base.

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CHEM 1011

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  1. CHEM 1011 pH and Buffer Solutions

  2. pH and Buffer Solutions Brønsted-Lowry Theory • Acid-proton donor • Base-proton acceptor

  3. pH and Buffer Solutions For example, HCl + H2O ↔ H3O+ + Cl- For the forward reaction, • HCl – proton donor - acid • H2O – proton acceptor - base

  4. pH and Buffer Solutions For example, HCl + H2O ↔ H3O+ + Cl- For the reverse reaction, • H3O+ – proton donor - acid • Cl- – proton acceptor - base

  5. pH and Buffer Solutions According to Brønsted-Lowry Theory, every acid-base reaction creates conjugate acid-base pair.

  6. pH and Buffer Solutions For example, HCl + H2O ↔ H3O+ + Cl- HCl - acid, Cl- - conjugate base H2O - base, H3O+ - conjugate acid

  7. pH and Buffer Solutions Acids can be monoprotic, diprotic, or triprotic.

  8. pH and Buffer Solutions • Monoprotic – donate one proton e.g., HCl, HNO3, HCOOH, CH3COOH HCl + H2O ↔ H3O+ + Cl- HNO3 + H2O ↔ H3O+ + NO3- HCOOH + H2O ↔ H3O+ + HCOO- CH3COOH + H2O ↔ H3O+ + CH3COO-

  9. pH and Buffer Solutions • Diprotic – donate two protons e.g, H2SO4, H2CO3 H2SO4 + H2O ↔ H3O+ + HSO4- HSO4- + H2O ↔ H3O+ + SO42- H2CO3 + H2O ↔ H3O+ + HCO3- HCO3- + H2O ↔ H3O+ + CO32-

  10. pH and Buffer Solutions • Triprotic – donate three protons e.g, H3PO4 H3PO4 + H2O ↔ H3O+ + H2PO4- H2PO4- + H2O ↔ H3O+ + HPO42- HPO42- + H2O ↔ H3O+ + PO43-

  11. pH and Buffer Solutions Water can self ionize. H2O + H2O ↔ H3O+ + OH- [H3O+] = 1 x 10-7 moles/L [OH-] = 1 x 10-7 moles/L Kw = [H3O+] [OH-] = [1 x 10-7] [1 x 10-7] = 1 x 10-14

  12. pH and Buffer Solutions • Kw same for all aqueous solutions. • Can calculate [OH-] for solution of strong acid. • e.g., 0.05 M HNO3 [H3O+] = 5 x 10-2 moles/L [OH-] = Kw / [H3O+] [OH-] = 1 x 10-14 / 5 x 10-2 = 2 x 10-13 mol/L

  13. pH and Buffer Solutions • Can also calculate [H3O+] of solution of strong base. • e.g., 0.04 M NaOH [OH-] = 4 x 10-2 mol/L [H3O+] = Kw / [OH-] [H3O+] = 1 x 10-14 / 4 x 10-2 = 2.5 x 10-13

  14. pH and Buffer Solutions • pH scale goes from 0 to 14 • pH < 7 acidic • pH = 7 neutral • pH > 7 basic

  15. pH and Buffer Solutions • pH = - log [H3O+] • e.g., pH of 0.05 M HNO3 (a strong acid) pH = - log [5 x 10-2] = 1.12

  16. pH and Buffer Solutions • The pOH of the solution of a strong base can also be calculated and used to determine pH. • e.g., 0.04 M NaOH pOH = -log[OH-] = -log[4 x 10-2] = 1.10 pH + pOH = pKw = 14 pH = 14 - pOH = 14 - 1.10 = 12.9

  17. pH and Buffer Solutions • You will use pH paper to measure the pHs of your solutions. Although a pH meter would yield more accurate results, it is unnecessary.

  18. pH and Buffer Solutions Buffer solutions • Resist changes of pH following the addition of acid or base. • Combinations of weak acids and their conjugate bases.

  19. pH and Buffer Solutions • Buffered aspirin • Blood buffers prevent acidosis and alkalosis. H2CO3 + H2O ↔ HCO3- + H3O+ H2PO4- + H2O ↔ HPO42- + H3O+

  20. pH and Buffer Solutions • The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer system. pH = pKa + log [A-] / [HA] • pH = pKa when [A-] = [HA] • Buffer activity continues as long as log [A-] / [HA] ratio kept constant

  21. Conclusions and Calculations • Answer Post-Lab questions 1 and 3 based upon your observations (omit 2).

  22. Conclusions and Calculations • For Post-Lab question 4, the HCl reacts with the CH3COO-, and the NaOH reacts with the H3O+ • An acetic acid-acetate buffer system would be represented by CH3COOH + H2O ↔ CH3COO- + H3O+

  23. Conclusions and Calculations • For an acetic acid-acetate buffer system pH = pKa + log [CH3COO-] / [CH3COOH] pKa = [CH3COO-][H3O+] / [CH3COOH] = 4.75 pH = 4.75 + log [CH3COO-] / [CH3COOH] • Use this relationship when answering Pre-Lab question 5 and Post-Lab questions 5 a. and b.

  24. Conclusions and Calculations • For Post-Lab questions 5 c. and d., H2CO3 + H2O ↔ HCO3- + H3O+ pH = pKa + log [HCO3-] / [H2CO3] pKa = [HCO3-][H3O+] / [H2CO3] = 6.37 pH = 6.37 + log [HCO3-] / [H2CO3]

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