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ACIDS AND BASES. REVISE pH = -log [H + ] pOH = -log [OH - ] pH + pOH = 14 at 25 o C Neutral:pH = 7([H + ] = [OH - ]) Acidic:pH [OH - ]) Basic:pH > 7([H.

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Presentation on theme: "ACIDS AND BASES. REVISE pH = -log [H + ] pOH = -log [OH - ] pH + pOH = 14 at 25 o C Neutral:pH = 7([H + ] = [OH - ]) Acidic:pH [OH - ]) Basic:pH > 7([H."— Presentation transcript:

1 ACIDS AND BASES

2 REVISE pH = -log [H + ] pOH = -log [OH - ] pH + pOH = 14 at 25 o C Neutral:pH = 7([H + ] = [OH - ]) Acidic:pH [OH - ]) Basic:pH > 7([H + ] < [OH - ]) Brønsted-Lowry acids and bases Brønsted-Lowry acids and bases Amphoteric substances Conjugate acid base pairs Neutralisation K w = [OH - ][H + ] K w = 14.0 at 25 o C K w = K a K b

3 STRONG ACIDS AND BASES Common strong acids: HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 4 (Why is HF not a strong acid?) Common strong bases: LiOH, NaOH, KOH, RbOH, CsOH, R 4 NOH Strong acids and bases  react nearly “completely” to produce H + and OH -  equilibrium constants are large e.g.: HCl H + + Cl - Complete dissociation: large small

4 Example: Calculate the pH of 0.1M LiOH. Strong base  Dissociates completely LiOH  Li + + OH - Start: Complete rxn: 0.1M00 0 pOH = -log [OH - ] = -log (0.1) = 1  pH = 14 - 1 = 13

5 Problem: What is the pH of 1x10 -8 M KOH? As before: pOH = -log (1x10 -8 ) = 8 pH = 14 – 8 = 6 BUT pH 6  acidic conditions and KOH is a strong base IMPOSSIBLE!!!!!!

6 Since the concentration of KOH is so low (1x10 -8 M), we need to take the ionisation of water into account. We do this by systematic treatment of equilibrium. Charge balance:[K + ] + [H + ] = [OH - ] Mass balance:[K + ] = 1x10 -8 M Equilibria:[H + ][OH - ] = K w = 1x10 -14 M 3 equations + 3 unknowns  solve simultaneously Find: pH = 7.02 In pure water [OH - ] = 1x10 -7 M, which is greater than the concentration of OH - from KOH. Hint: You end up with a quadratic equation which you solve using the formula.

7 Also note that: Only pure water produces 1x10 -7 M H + and OH -. If there is say 1x10 -4 M HBr in solution, pH = 4 and [OH - ] = 1x10 -10 M But the only source of OH - is from the dissociation of water.  if water produces 1x10 -10 M OH -  it can only produce 1x10 -10 M H + due to the dissociation of water.  pH in this case is due mainly to the dissociation of HBr and not the dissociation of water. It is thus important to look at the concentration of acid and bases present.

8 Some guidelines regarding the concentrations of acids and bases: 1)When conc > 1x10 -6 M  calculate pH as usual 2)When conc < 1x10 -8 M  pH = 7 (there is not enough acid or base to affect the pH of water) 3)When conc  1x10 -8 - 1x10 -6 M  Effect of water ionisation and added acid and bases are comparable, thus: use the systematic treatment of equilibrium approach.

9 WEAK ACIDS AND BASES HA H + + A - Acid dissociation constant Partial dissociation small large Weak acids and bases  react only “partially” to produce H + and OH -  equilibrium constants are small Common weak acids: carboxylic acids (e.g. acetic acid = CH 3 COOH) ammonium ions (e.g. RNH 3 +, R 2 NH 2 +, R 3 NH + ) Common weak bases: carboxylate anions (e.g. acetate = CH 3 COO - ) amines (e.g. RNH 2, R 2 NH, R 3 N)

10 Base hydrolysis: B + H 2 O BH + + OH - base hydrolysis constant/ base “dissociation” constant Weak base  partial dissociation  K b small NOTE: pK a = -log K a pK b = -log K b As K increases, its p-function decreases and vice versa.

11 Problem: Find the pH of a solution of formic acid given that the formal concentration is 2 M and K a = 1.80x10 -4. Systematic treatment of equilibria: Charge balance:[H + ] = [HCOO - ] + [OH - ] Mass balance:2 M = [HCOOH] + [HCOO - ] Equilibria: HCOOH H + + HCOO - H 2 O H + + OH - 4 equations  4 unknowns  difficult to solve

12  Make an assumption: [H + ] due to acid dissociation  [H + ] due to water dissociation Produces HCOO - [HCOO - ] large Produces OH - [OH - ] small  [HCOO - ] >> [OH - ]  Charge balance:[H + ]  [HCOO - ]

13  [H + ] = [HCOO - ] = 0.019 M Charge balance: [H + ]  [HCOO - ] Mass balance:2 M = [HCOOH] + [H + ] Equilibria: Let [H + ] = [HCOO - ] = x Or x = -0.019  No negative conc’s  pH = 1.7

14 OR since [HCOOH] > 1x10 -6, we can calculate pH as usual HCOOH H + + HCOO - Start: Equilibrium: 2M00 2-xxx Weak acid  equilibrium conditions Solve as before

15 FRACTION OF DISSOCIATION,  Fraction of acid in the form A - For the above problem:  = [HCOO - ] F = 0.019 M 2 M = 0.0095  Acid is 0.95% dissociated at 2 M formal concentration Weak electrolytes dissociate more as they are diluted.

16 WEAK BASE EQUILIBRIA Charge balance:[BH + ] = [OH - ] Mass balance:F = [B] + [BH + ] Equilibria: Let [BH + ] = [OH - ] = x B + H 2 O BH + + OH - FRACTION OF ASSOCIATION

17 Relationship between K a and K b for a conjugate acid- base pair: K a.K b = K w = 1x10 -14 at 25 o C CONJUGATE ACIDS AND BASES  If K a is very large (strong acid) Then K b must be very small (weak conjugate base) Base so weak it is not a base at all in water And vice versa If K a is very small, say 1x10 -6 (weak acid) Then K b must be small, 1x10 -8 (weak conjugate base) Greater acid strength, weaker conjugate base strength, and vice versa.

18 Problem: Calculate the pH of 0.1 M NH 3, given that pK a = 9.244 for ammonia. NH 3 + H 2 O NH 4 + + OH - acidbase KbKb pK a = -log K a  K a = 5.70x10 -10 = 1.75 x10 -5 K b = KwKw KaKa = 1x10 -14 5.70x10 -10

19 Problem: Calculate the pH of 0.1 M NH 3, given that pK a = 9.244 for ammonia. = 1.75 x10 -5 K b = x2x2 F - x = x2x2 0.1 - x Solve for x using the quadratic equation Where x = [OH - ] = [NH + ] Find x = 1.31x10 -3 M = [OH - ] pOH = -log [OH - ]= 2.88  pH = 14 – 2.88 = 11.12 Negative value discarded

20 BUFFERS Mixture of an acid and its conjugate base. Buffer solution  resists change in pH when acids or bases are added or when dilution occurs. Mix: A moles of weak acid + B moles of conjugate base Find: moles of acid remains close to A, and moles of base remains close to B  Very little reaction HA H + + A - Le Chatelier’s principle

21 HENDERSON-HASSELBALCH EQUATION For acids: For bases: B + H 2 O BH + + OH - acidbaseacidbase   KaKa KbKb pK a applies to this acid When [A - ] = [HA], pH = pK a

22 HA H + + A - Derivation:

23 Why does a buffer resist change in pH when small amounts of strong acid or bases is added? The acid or base is consumed by A - or HA respectively A buffer has a maximum capacity to resist change to pH. Buffer capacity,  :  Measure of how well solution resists change in pH when strong acid/base is added. ? Larger   more resistance to pH change

24 A buffer is most effective in resisting changes in pH when: pH = pK a i.e.:[HA] = [A - ]  Choose buffer whose pK a is as close as possible to the desired pH. pK a  1 pH unit

25 Problem: Calculate the pH of a solution containing 0.200 M NH 3 and 0.300 M NH 4 Cl given that the acid dissociation constant for NH 4 + is 5.7x10 -10. NH 3 + H 2 O NH 4 + + OH - acidbase KaKa pK a applies to this acid pK a = 9.244 pH = 9.244 + log (0.200) (0.300) pH = 9.07

26 POLYPROTIC ACIDS AND BASES Can donate or accept more than one proton. Relationships between K a ’s and K b ’s: K a1. K b2 = K w K a2. K b1 = K w Diprotic acid: H 2 L HL - + H + K a1  K 1 HL - L 2- + H + K a2  K 2 Diprotic base: L 2- + H 2 O HL - + OH - K b1 HL - + H 2 O H 2 L + OH - K b2 In general:

27 Using pK a values and mass balance equations, the fraction of each species can be determined at a given pH.

28

29 ACID-BASE TITRATIONS We will construct graphs to see how pH changes as titrant is added. Start by: writing chemical reaction between titrant and analyte using the reaction to calculate the composition and pH after each addition of titrant

30 TITRATION OF STRONG BASE WITH STRONG ACID Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr. HBr + KOH KBr + H 2 O What is of interest to us in an acid-base titration: H + + OH - H 2 O Mix strong acid and strong base  reaction goes to completion H + + OH -  H 2 O

31 * Calculate volume of HBr needed to reach the equivalence point, V eq : C1V1C1V1 C2V2C2V2 n1n1 n2n2 = But n 1 = n 2 = 1  C HBr V eq = C KOH V KOH (0.1000 M)V eq = (0.02000 M)(50.00 ml) V eq = 10.00 ml Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.

32 There are 3 parts to the titration curve: 1)Before reaching the equivalence point  excess OH - present 2)At the equivalence point  [H + ] = [OH - ] 3)After reaching the equivalence point  excess H + present 1 2 3

33 1)Before reaching the equivalence point  excess OH - present Say 2.00 ml HBr has been added. C OH- = n unreacted V total Starting n OH- = (0.02 M)(0.050 L) = 1x10 -3 mol n H+ added = (0.1 M)(0.002 L) = 2x10 -4 mol  n OH- unreacted = 8x10 -4 mol V total = 50 + 2 mL = 52 mL = 0.052 L C OH- = 0.01538 M HBr + KOH KBr + H 2 O K w = [H + ][OH - ] 1x10 -14 = [H + ](0.01538 M) [H + ] = 6.500x10 -13 M  pH = 12.19

34 2)At the equivalence point  n H+ = n OH- pH is determined by dissociation of H 2 O: H 2 O H + + OH - xx K w = [H + ][OH - ] 1x10 -14 = x 2 x = 1x10 -7 M  [H + ] = 1x10 -7 M  pH = 7 pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!

35 3)After reaching the equivalence point  excess H + present Say 10.10 ml HBr has been added. C H+ = n excess V total Starting n OH- = 1x10 -3 mol n H+ added = (0.1 M)(0.0101 L) = 1.010x10 -3 mol  n H+ excess = 1x10 -5 mol V total = 50 + 10.1 mL = 60.1 mL = 0.0601 L C H+ = 1.664x10 -4 M pH = 3.78 HBr + KOH KBr + H 2 O

36 Note: A rapid change in pH near the equivalence point occurs. Equivalence point where: slope is greatest second derivative is zero (point of inflection)

37 Calculate titration curve by calculating pH values after a number of additions of HBr.

38 TITRATION OF WEAK ACID WITH STRONG BASE Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO 2 H + NaOH HCO 2 Na + H 2 O HA A-A- OR HCO 2 H + OH - HCO 2 - + H 2 O pK a = 3.745 K a = 1.80x10 -4 K b = 5.56x10 -11 Equilibrium constant so large  reaction “goes to completion” after each addition of OH - Strong and weak react completely

39 * Calculate volume of NaOH needed to reach the equivalence point, V eq : C1V1C1V1 C2V2C2V2 n1n1 n2n2 = But n 1 = n 2 = 1  C NaOH V eq = C FA V FA (0.1000 M)V eq = (0.02000 M)(50.00 ml) V eq = 10.00 ml Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO 2 H + OH - HCO 2 - + H 2 O

40 There are 4 parts to the titration curve: 2)From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A - HA + OH - A - + H 2 O BUFFER!!  use Henderson-Hasselbalch eqn for pH 1)Before base is added  HA and H 2 O present. HA weak acid,  pH determined by equilibrium: HA H + + A - KaKa 1 2

41 4) Beyond the equivalence point  excess OH - added to A -. Good approx: pH determined by strong base (neglect small effect from A - ) 3)At the equivalence point  all HA converted to A -. A - is a weak base whose pH is determined by reaction: A - + H 2 O HA + OH - KbKb 3 4

42 1)Before base is added  HA and H 2 O present. HA = weak acid. x 2 + 1.80x10 -4 x – 3.60x10 -6 = 0 x = 1.81x10 -3  [H + ] = 1.81x10 -3  pH = 2.47 K a = 1.80x10 -4 HA H + + A - F- x xx 1.80x10 -4 = x2x2 0.02 - x

43 2)From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A -. BUFFER!! pH = 3.745 + log 2x10 -4 8x10 -4 Say 2.00 ml NaOH has been added. HA + OH - A - + H 2 O  pH = 3.14 Starting n HA = (0.02 M)(0.05 L) = 1x10 -3 mol n OH- added = (0.1 M)(0.002 L) = 2x10 -4 mol 1x10 -3 2x10 -4 8x10 -4 2x10 -4 Start - - End HCO 2 H + OH - HCO 2 - + H 2 O

44 But V A = V HA = V Tot

45 When volume of titrant = ½ V eq pH = pK a Special condition: n HA = n A- Since:

46 3)At the equivalence point  all HA converted to A -. A - = weak base. (n HA = n NaOH ) Starting n HA = 1x10 -3 mol  Solution contains just A -  a solution of weak base HA + OH - A - + H 2 O 1x10 -3 Start - - End -  n OH- = 1x10 -3 mol

47 A - + H 2 O HA + OH - F- x xx K b = 5.56x10 -11 = 0.0167 M V total = 50 + 10 mL = 60 mL = 0.060 L x 2 + 5.56x10 -11 x – 9.27x10 -13 = 0 x = 9.63x10 -7 pH = 7.98 [OH - ] = 9.63x10 -7 M pOH = 6.02 pH is slightly basic at equivalence point for strong base-weak acid titrations F A- = nA-nA- V = 1x10 -3 mol 0.060 L 5.56x10 -11 = x2x2 0.0167 - x

48 CALCULATED TITRATION CURVE

49 Titration curve depends on K a of HA. As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate an acid or base that is too weak.

50 Titration curve depends on extent of dilution of HA. As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate a very dilute acid or base.

51 TITRATION OF WEAK BASE WITH STRONG ACID This is the reverse of the titration of weak base with strong acid. The titration reaction is: B + H + BH + Recall: Strong and weak react completely

52 There are 4 parts to the titration curve: 1)Before acid is added  B and H 2 O present. B weak base  pH determined by equilibrium: B + H 2 O BH + + OH - KbKb F- x xx

53 2)From first addition of acid to immediately before equivalence point  mixture of unreacted B and BH + B + H + BH + BUFFER!!  use Henderson-Hasselbalch equation for pH pK a applies to this acid

54 3)At the equivalence point  all B converted to BH +. BH + is a weak acid  determined pH by reaction: BH + B + H + Ka F-xF-x xx F BH+ = n V total Take dilution into account pH is slightly acidic (pH below 7) for strong acid-weak base titrations

55 4) Beyond the equivalence point  excess H + added to BH +. Good approx: pH determined by strong acid (neglect small effect from BH + ) Example: 50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. K a for NaCN = 6.20x10 10 Draw the titration curve by calculating pH at various volumes of HCl.

56 TITRATION CURVE OF WEAK BASE WITH STRONG ACID

57 TITRATIONS IN DIPROTIC SYSTEMS Example - a base that is dibasic: B + H +  BH + BH + + H +  BH 2 2+ pK b1 = 4.00 pK b2 = 9.00 With corresponding reactions: Two end points will be observed.

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59 FINDING END POINTS WITH A pH ELECTRODE After each small addition of titrant the pH is recorded and a titration curve is plotted. 2 ways of determining end points from this: using derivatives using a Gran plot

60 Setup

61 But there are autotitrators! Titrando from Metrohm

62 USING DERIVATIVES End point is taken where the slope is greatest Or where the 2 nd derivative is zero

63 USING A GRAN PLOT A problem with using derivatives  titration data is most to obtain near the end point Example – titration of a weak acid, HA HA H + + A - K a = [H + ]  H+ [A - ]  A- [HA]  HA NOTE: pH electrode responds to hydrogen ion ACTIVITY, not concentration

64 Say we titrated V a ml of HA (formal conc = F a ) with V b ml of NaOH (formal conc = F b ) : HA + OH - A - + H 2 O [A - ] = n OH (titrated) V Total = VbFbVbFb V a + V b [HA] = n HA (initial) – n OH (titrated) V Total = V a F a - V b F b V a + V b Substitute into the equilibrium constant: K a = [H + ]  H+ [A - ]  A- [HA]  HA

65 Rearrange: [H + ]  H+ = 10 -pH = V e - V b VaFaVaFa FbFb - V b

66 Gran plot equation: Gran plot  Graph of V b 10 -pH vs V b If is constant, then:  A-  HA Slope = -K a  A-  HA andx-intercept = V e Use data taken before end point to find end point Can determine K a from slope

67 Use only linear portion of graph Extrapolate graph to get V e

68 FINDING END POINTS WITH INDICATORS Acid-base indicator  acid or base itself Various protonated species have different colours HIn H + + In -

69 Choose indicator whose colour change is as close as possible to the pH of the end point Indicators transition range overlaps the steepest part of the titration curve

70 Indicator error: difference between the observed end point (colour change) and the true equivalence point. Systematic error Random error Visual uncertainty associated with distinguishing the colour of the indicator reproducibly Why do we only add a few drops of indicator? Indicator is an acid/base itself  will react with analyte/titrant Few drops  neglible relative to amount of analyte

71 Good luck with the test and don’t panic!

72 HOW LONG DOES IT TAKE YOU TO FIND THE MAN’S HEAD?


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