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Acids and bases, pH and buffers

Acids and bases, pH and buffers. Dr. Mamoun Ahram Lecture 2. Acids and bases. Acids versus bases. Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4) Base: a substance that produces OH- when dissolved in water (NaOH, KOH) What about ammonia (NH3)?.

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Acids and bases, pH and buffers

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  1. Acids and bases, pH and buffers Dr. Mamoun Ahram Lecture 2

  2. Acids and bases

  3. Acids versus bases • Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4) • Base: a substance that produces OH- when dissolved in water (NaOH, KOH) • What about ammonia (NH3)?

  4. Brønsted-Lowry acids and bases • The Brønsted-Lowry acid: any substance able to give a hydrogen ion (H+-a proton) to another molecule • Monoprotic acid: HCl, HNO3, CH3COOH • Diprotic acid: H2SO4 • Triprotic acid: H3PO3 • Brønsted-Lowry base: any substance that accepts a proton (H+) from an acid • NaOH, NH3, KOH

  5. Acid-base reactions • A proton is transferred from one substance (acid) to another molecule Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A- • Ammonia is base • HA is acid • Ammonium ion (NH4+) is conjuagte acid • A- is conjugate base

  6. Water: acid or base? • Both • Products: hydronium ion (H3O+) and hydroxide

  7. Amphoteric substances • Example: water NH3 (g) + H2O(l) ↔ NH4+(aq) + OH–(aq) HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

  8. Acid-base reactions Acid + base  salt + H2O • Exceptions: • Carbonic acid (H2CO3)-Bicarbobate ion (HCO3-) • Ammonia (NH3)-

  9. Acid/base strength

  10. Rule • The stronger the acid, the weaker the conjugate base HCl(aq)→ H+(aq) + Cl-(aq) NaOH(aq)→ Na+(aq) + OH-(aq) HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq) NH3 (aq) + H2O(l)↔ NH4+(aq) + OH-(aq)

  11. Equilibrium constant    HA  <-->   H+ + A- Ka: >1 vs. <1

  12. Expression • Molarity (M) • Normality (N) • Equivalence (N)

  13. Molarity of solutions moles = grams / MW M = moles / volume (L) grams = M x vol (L) x MW

  14. Exercise • How many grams do you need to make 5M NaCl solution in 100 ml (MW 58.4)? grams = 58.4 x 5 moles x 0.1 liter = 29.29 g

  15. Normal solutions N= n x M (where n is an integer) • n =the number of donated H+ Remember! The normality of a solution is NEVER less than the molarity

  16. Equivalents • The amount of molar mass (g) of hydrogen ions that an acid will donate • or a base will accept • 1M HCl = 1M [H+] = 1 equivalent • 1M H2SO4 = 2M [H+] = 2 equivalents

  17. Exercise • What is the normality of H2SO3 solution made by dissolving 6.5 g into 200 mL? (MW = 98)?

  18. Example One equivalent of Na+ = 23.1 g One equivalent of Cl- - 35.5 g One equivalent of Mg+2 = (24.3)/2 = 12.15 g Howework: Calculate milligrams of Ca+2 in blood if total concentration of Ca+2 is 5 mEq/L.

  19. Titration • The concentration of acids and bases can be determined by titration

  20. Excercise • A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl? • Step 1 - Determine [OH-] • Step 2 - Determine the number of moles of OH- • Step 3 - Determine the number of moles of H+ • Step 4 - Determine concentration of HCl

  21. A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl • Moles of base = Molarity x Volume • Moles base = moles of acid • Molarity of acid= moles/volume

  22. Another method MacidVacid = MbaseVbase

  23. Note • What if one mole of acid produces two moles of H+ MacidVacid = 2MbaseVbase

  24. Homework • If 19.1 mL of 0.118 M HCl is required to neutralize 25.00 mL of a sodium hydroxide solution, what is the molarity of the sodium hydroxide? • If 12.0 mL of 1.34 M NaOH is required to neutralize 25.00 mL of a sulfuric acid, H2SO4, solution, what is the molarity of the sulfuric acid?

  25. Equivalence point

  26. Ionization of water H3O+ = H+

  27. Equilibrium constant Keq = 1.8 x 10-16 M

  28. Kw • Kw is called the ion product for water

  29. pH

  30. What is pH?

  31. Acid dissociation constant • Strong acid • Strong bases • Weak acid • Weak bases

  32. pKa

  33. What is pKa?

  34. Henderson-Hasselbalch equation

  35. The equation pKa is the pH where 50% of acid is dissociated into conjugate base

  36. Buffers

  37. Maintenance of equilibrium

  38. What is buffer?

  39. Titration

  40. Midpoint

  41. Buffering capacity

  42. Conjugate bases

  43. How do we choose a buffer?

  44. Problems and solutions • A solution of 0.1 M acetic acid and 0.2 M acetate ion. The pKa of acetic acid is 4.8. Hence, the pH of the solution is given by • Similarly, the pKa of an acid can be calculated

  45. Exercise • What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4)

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