2. Acids & Bases and Titrations

3. Properties of Acids & Bases Acids –taste sour –feel like water –good electrolytes –turn blue litmus paper “pink” –pH less than 7 at 25°C Bases –taste bitter –feel slippery –good electrolytes –turn pink litmus paper “blue” –pH greater than 7 at 25°C

4. 3 Definitions of Acids Arrhenius: acids are substances which produce H + ions in water. Brønsted-Lowry: acids are substances which donate H + ions to other substances. Lewis: acids are substances which accept electron pairs from other substances.

5. 3 Definitions of Bases Arrhenius: Bases are substances that produce OH - ions in water. Brønsted-Lowry: Bases are substances that accept H + ions from other substances. Lewis: Bases are substances that donate a pair of electrons to another substance.

6. Conjugate Acids-Base Pairs HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) acid base acid base SO 4 2- (aq) + HCO 3 - (aq)  HSO 4 - (aq) + CO 3 2- (aq) base acid acid base

8. Acid Strength When an acid is placed in water, it dissociates as it donates its H + ions to water molecules: HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) A strong acid is one which completely dissociates. This means it has a LARGE ionization constant, K a. (the equilibrium constant for this reaction) It is a very good electrolyte.

9. Relative Acid Strength

10. Hydrochloric acid, HCl Nitric acid, HNO 3 Hydrobromic acid, HBr Hydroiodic acid, HI Sulfuric acid, H 2 SO 4 (only the first H + dissociates strongly) Common Strong Acids

11. Common Weak Acids Hydrofluoric acid, HF K a = 6.8 x 10 -4 Acetic acid, CH 3 COOH K a = 1.8 x 10 -5 Nitrous acid, HNO 2 K a = 4.5 x 10 -4 Benzoic acid, C 6 H 5 COOH K a = 6.5 x 10 -5 Rank these acids in order of increasing strength. Write a dissociation reaction for each acid. Write the K a expression for each acid.

13. Which of these "molecular" pictures best represents a concentrated solution of the weak acid HA, with Ka = 1 x 10 -5 ? Explain your choice. Weak Acids

14. Polyprotic Acids A polyprotic acid is one that has more than one H + to donate: Monoprotic acids: HF, HCl, HNO 3 Diprotic acids: H 2 SO 4, H 2 CO 3 Triprotic acid: H 3 PO 4 Polyprotic acids dissociate “stepwise”: e.g. Sulfuric Acid: H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - HSO 4 - + H 2 O  H 3 O + + SO 4 2- Write the 3 steps in the dissociation of phosphoric acid! Write the Ka expression for each step!

15. Dissociation Constants - Polyprotic Acids

16. Amphoteric Substances An amphoteric substance is a substance which can behave as either an acid or a base. eg. HNO 3 (aq) + H 2 O (l)  H 3 O + (aq) + NO 3 - (aq) NH 3 (aq) + H 2 O (aq)  NH 4 + (aq) + OH - (aq) To be amphoteric, a substance must have a hydrogen ion to donate. Most amphoteric substances are negatively charged. (e.g. HCO 3 -, H 2 PO 4 - etc…)

17. The pH Scale Acids produce H + ions in water that immediately react with water to form hydronium ions, H 3 O +. The concentration of H + (or H 3 O + ) in a solution is a measure of the solution’s acidity. pH is one way to express the hydronium concentration. pH = - log [H 3 O + ] W hen [H 3 O + ] = 0.010 M in a solution, the pH = 2.00. When the pH of a solution is 8.00, [H 3 O + ] = 1.0 x 10 -8 M. What is [H 3 O + ] when the pH = 3.25?

18. The pH Scale (continued)

19. Acidic, Basic, or Neutral?

20. Auto-Ionization of Water We’ve seen that water is amphoteric. In pure water, a very small fraction of the water molecules undergo an acid-base reaction with each other: H 2 O + H 2 O  H 3 O + + OH - At 25°C, the equilibrium constant (K w ) for this reaction is very small. K w = 1.0 x 10 -14 =[H 3 O + ][OH - ] A simple equilibrium calculation shows that in pure water at 25°C, [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M. Thus, at 25°C, the pH of pure water is 7.00 (and pOH = 7.00).

21. Ion product constant @ 25 o C: K w =[H 3 O + ][OH - ]= 1.0 X 10 -14

22. A Logarithm Lesson K w = 1.0 x 10 -14 =[H 3 O + ][OH - ] - log (1.0 x 10 -14 ) = - log ([H 3 O + ][OH - ]) 14.00 = -log [H 3 O + ] + -log[OH - ] 14.00 = pH + pOH

24. pH Practice

25. Calculate the pH of 0.10-M HNO 3 solution. HNO 3 + H 2 O  H 3 O + + NO 3 - 0.10 M ---- ---- - 0.10 + 0.10 + 0.10 ---- 0.10 M 0.10 M Thus, [H 3 O + ] = 0.10 M pH = - log (0.010) = 2.00 HNO 3 is a strong acid - Ka is large! It dissociates completely. Why do we let [H 3 O + ] = 0 at the start? Isn’t there some H 3 O + in pure water?? Are there any HNO 3 molecules in the solution?? Would this solution be a good electrolyte??

26. Calculate the pH of 0.15-M HF solution. HF + H 2 O  H 3 O + + F - 0.15 M -------- - x + x + x 0.15 - x x x Since Ka is small let’s assume that x << 0.15 M x 1 = 0.010 M x 2 = 0.0097 M x 3 = 0.0097 M = [H 3 O + ] pH = - log (0.0097) = 2.01 HF is a weak acid - its Ka is 6.7 x 10 -4. It doesn’t dissociate completely.

27. Percent Dissocation (Ionization) Calculate the %-Dissociation of the HF molecules in the 0.15-M solution of HF. Of the 0.15 mol/L HF molecules, we know that 0.0097 mol/L have dissociated. Thus… % Dissociation = = 6.5% Does it make sense that the %-dissociation is so small for HF?

28. % Dissociation - Effect of Concentration

30. Calculate the pH of 0.20M H 2 SO 4 (aq) H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - 0.20 M ---- ---- - 0.20 + 0.20 + 0.20 ---- 0.20 M 0.20 M H 2 SO 4 is a diprotic acid - it dissociates in 2 steps. Ka for the first dissociation is LARGE. HSO 4 - + H 2 O  H 3 O + + SO 4 2- 0.20 M 0.20 M ---- - x + x + x 0.20 - x 0.20 + x x Note that the second step starts with both HSO 4 - and H 3 O + at 0.20 M. The second dissociation is weak - Ka 2 = 0.012.

31. pH of 0.20M H 2 SO 4 (aq) (cont’d) Since the Ka is small for the second dissociation, we’ll assume that x << 0.20 M Note that the assumption applies TWICE - in the numerator and denominator. x 2 = 0.011 M This is the [H 3 O + ] from the 2nd dissociation. [H 3 O + ] for both steps = 0.20 + 0.011 = 0.21 M pH = - log (0.21) = 0.68 Find the TOTAL [H 3 O + ] from BOTH steps of the dissociation in order to find the pH.

32. Other Polyprotic Acids For most polyprotic acids (H 2 CO 3, H 3 PO 4 etc…) the first dissociation is the only one that contributes significantly to the [H 3 O + ] of the solution. This is because Ka 1 >> Ka 2, Ka 3 Thus, the first step is the only one that needs to be considered and the acids can be treated as weak monoprotic acids.

33. Find the pH of 0.10 M H 3 PO 4 H 3 PO 4 + H 2 O  H 3 O + + H 2 PO 4 2- 0.10 M ---- ---- - x + x + x 0.10 - x x x H 3 PO 4 is a triprotic acid - it dissociates in 3 steps. Ka for the first dissociation is small. Ka 1 = 7.5 x 10 -3 Since Ka is small assume that x << 0.10 M x 1 = 0.027 M x 2 = 0.023 M x 3 = 0.024 M = [H 3 O + ] Since Ka 1 >> Ka 2, Ka 3 assume these steps are negligible pH = - log (0.024) = 1.62

34. Strong Bases Strong bases are metal hydroxides: NaOH  Na + + OH - KOH  K + + OH - LiOH  Li + + OH - Ca(OH) 2  Ca 2+ + 2 OH - Sr(OH) 2  Sr 2+ + 2 OH - “Strong” refers to the dissociation of the bases. Strong bases dissociate completely into ions. They are good electrolytes Notice that hydroxide ions are produced in a basic solution

36. pH of Strong Base Solutions Calculating the pH of a strong base solution is relatively easy because strong bases completely dissociate in water. For example, calculate the pH of 0.010-M NaOH solution. NaOH  Na + + OH - In a 0.010-M solution, [Na + ] = [OH - ] = 0.010 M Thus, pOH = - log [OH - ] = - log (0.010) = 2.00 And pH = 14.00 - pOH = 14.00 - 2.00 = 12.00 Strong Bases dissociate 100 %

37. Weak Bases As you can see, most of the weak bases that are commonly encountered are derived from ammonia, NH 3. Note that in addition to the weak bases listed below, every weak acid has a corresponding weak base.

38. NH 3 + H 2 O  NH 4 + + OH - 0.25 M ---- ---- - x + x + x 0.25 - x x x Calculate the pH of 0.25-M NH 3 solution. Since K b is small let’s assume that x << 0.25 M x 1 = 0.0021 M x 2 = 0.0021 M = [OH - ] pOH = - log (0.0021) = 2.68 pH = 14.00 - pOH = 11.32 NH 3 is a weak base - its K b is 1.8 x 10 -5. It doesn’t dissociate completely.

39. Relationship between K a and K b Consider the dissociation of HF and its conjugate base in water: HF + H 2 O  H 3 O + + F - F - + H 2 O  HF + OH - K a · K b = K w

40. Relative Strengths of Conjugate Acid-Base Pairs

41. Acid-Base Properties of Ions

42. pH of Salt Solutions Salts are made of cations and anions. A few cations may be acidic (NH 4 + ) Many anions may be basic (F -, SO 4 2- etc…) Some ions are amphoteric (HCO 3 - ) Salt solutions will be acidic, basic or neutral depending on the ions that make up the salt. e.g. Sodium Chloride NaCl  Na + + Cl - Na + = neither acid nor base = neutral Cl - = conj. base of HCl (strong acid) = K b too small to affect pH = neutral Thus, NaCl is a neutral salt.

43. pH of Salt Solutions (cont’d) e.g. Ammonium nitrate, NH 4 NO 3 NH 4 + = conj. acid of ammonia, NH 3 (weak base) = ACIDIC NO 3 - = conj. base of nitric acid, HNO 3 (strong acid) = K b too small to affect pH = NEUTRAL Thus, NH 4 NO 3 is ACIDIC e.g. Sodium Bicarbonate NaHCO 3 Na + = NEUTRAL HCO 3 - = AMPHOTERIC Acid: Ka = 5.6 x 10 -11 Base: K b = K w ÷ K a (of H 2 CO 3 ) = 2.3 x 10 -8 Since its K b > K a, HCO 3 - will be BASIC Thus, NaHCO 3 is BASIC

44. Acid-Base Titrations A titration is a volumetric technique often involving acid-base neutralizations. It often involves reacting one solution of known concentration, with another of unknown concentration. Let’s take a look at how a titration is performed. Then we’ll look at calculations.

45. Titration Description A titration is a procedure for quantitative analysis. In a titration two reagents are mixed, one with a known concentration and one with an unknown concentration. An acid-base indicator is added to determine when the two reagents have reacted essentially completely At the end of the titration the unknown solution's concentration (or molar mass) can be calculated.

46. Description One reagent is a solution and is added from a buret. This solution is called the titrant. The solution from the buret is added to a flask that contains either a measured volume of a solution or a weighed quantity of solid that has been dissolved.

47. Preparing a Buret Rinse a clean buret three with ~ 5 mL portions of distilled water. Repeat the rinse using the titrant (the solution that will be added to the flask). Allow the rinse to drain through the buret stopcock to rinse the tip of the buret. Discard the rinse portions.

48. Fill the Buret Clamp the buret into place on the buret stand. Use a funnel to fill the buret with the titrant. Allow some titrant to drain out the tip. Remove any air bubbles in the stopcock tip.

49. Reading the Volume in the Buret Make sure the volume reading in the buret is at or below the 0.00 mL mark. Record the volume reading on the buret. Be sure to read the bottom of the meniscus. Record all certain digits plus one uncertain digit. suggest a better way to fill the buret!

50. Prepare the Sample If the sample to be titrated is a solution, pipet the desired volume into an Erlenmeyer flask. Record the exact volume transferred. Dilute the sample with a small portion of distilled water (about 10 to 20 mL). If the sample is a solid, obtain the desired mass, and dissolve it in about 25 mL of distilled water in an Erlenmeyer flask. Record the exact mass of sample used. A few drops of chemical indicator is usually added to signal the endpoint of the titration with a colour change. The choice of indicator depends on the pH at the end of the neutralization reaction (consider the salt produced!)

53. Add the Titrant At the beginning of the titration, titrant may be added quickly since the indicator color disappears rapidly. When the color persists for longer periods of time, add titrant more slowly (a drop or less at a time).

54. Adding the Titrant (cont’d) Be sure to mix the two reactant solutions thoroughly by swirling the flask as the titrant is added. If solution splashes up to the side of the flask, you can use distilled water to wash it back into the solution. Placing the flask on a piece of white paper will often help you observe the first appearance of color change.

55. Determining the Endpoint Overshooting the endpoint of the titration by adding too much titrant is a common error. The endpoint for this titration is reached when you reach a pale color that persists throughout the solution for several seconds. Record the volume in the buret.

56. Endpoint Problems Suggest what went wrong in each titration below!

57. Titration Calculations Step 1: Write a balanced equation for the neutralization reaction. Step 2: Write a mathematical equation using the mole-ratio for the acid and base. Step 3: For solutions, use “CV” to represent moles. For solids, use “m/M” to represent moles. Step 4: Substitute the given information into the equation. Step 5: Solve for the unknown quantity.

58. Titration Calculations - example A 25.00-mL sample of carbonic acid, H 2 CO 3, required 22.34 mL of 0.126-M potassium hydroxide, KOH, for complete neutralization. Calculate the concentration of the carbonic acid solution. H 2 CO 3 + 2 KOH  K 2 CO 3 + 2 H 2 O moles KOH = 2 (moles H 2 CO 3 ) C b V b = 2 C a V a

59. Titration Calculations - example An unknown monoprotic acid, HA, was titrated using 0.0981-M NaOH. A 0.214-g sample of the acid required 32.70 mL of the base for complete neutralization. Calculate the molar mass of the unknown acid. HA + NaOH  NaA + H 2 O moles HA = moles NaOH