Chemical Reactions: pH Equilibria
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Chemical Reactions: pH Equilibria

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Lecture materials for the Introductory Chemistry course for Forensic Scientists, University of Lincoln, UK. See http://forensicchemistry.lincoln.ac.uk/ for more details.

Lecture materials for the Introductory Chemistry course for Forensic Scientists, University of Lincoln, UK. See http://forensicchemistry.lincoln.ac.uk/ for more details.

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  • Strong acids dissociate completely in water. Concentrated acids contain several moles of substance per dm3.
  • When an acid dissociated, the H+ ions actually combine with H2O to produce H30+ (oxonium ions, or hydroxonium ions). H+ + H2O  H3O+ A- is referred to as the conjugate base of the acid; I.e.the ions associated with the acid and capable of recombining with the H+ ion. More on this later. E.g. Ka values HCl = 7.2 x 10 10 acetic acid = 1.8 x 10-5
  • Water can dissociate:I.e. separate into its ionic components. This dissociation is very partial, in fact one water molecule in 550 million is ionised at any given moment. Because of this to all intents and purposes the concentration of water is taken to be constant
  • Kw is really Keq x another constant for H2O.
  • Hopefully you have all used at some point universal pH paper green = 7 blue = very alkaline red = very acidic
  • 0.1M HNO3 is fully ionised: HNO3  H+ + NO3- thus, have 0.1moles in 1dm3 of H+ ions. Thus put 0.1 into equation to give a pH equalling 1. CH3COOH is partially ionised: CH3COOH  CH3C00- + H+ only have 0.001 or 10-3 moles of H+ ions, thus pH = 3. A pH meter musing a H electrode to measure the pH of a solution.
  • 0.1M NaOH - strong alkali-completely ionised NaOH  Na + + OH- Thus, 0.1M NaOH contains 0.1M of OH- ions in 1 dm3. Put into equation. 10-13 which gives a pH = 13.
  • Used in the lab: reagents can be added to solutions without pH altering and reactions carried out. Absorption of CO2 and lab fumes may also the pH of a solution. 15.7 – Housecroft – P462 - 467 Making use of the partial dissociation of weak acids and bases
  • P462
  • P466 housecroft

Chemical Reactions: pH Equilibria Chemical Reactions: pH Equilibria Presentation Transcript

  • This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Reversible Reactions and Chemical Equilibrium University of Lincoln presentation
  • Outline
    • Reversible reactions
    • Chemical Equilibrium
    • Le Chatelier’s Principle
    • Equilibrium constants
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  • Reversible Reactions
    • BiCl 3 (aq) + H 2 O(l) ↔ BiOCl(s) + 2HCl(aq)
    • CH 3 CO 2 H + CH 3 CH 2 OH ↔ CH 3 CO 2 CH 2 CH 3 + H 2 O
    • Cr 2 O 7 2- (aq) + 2OH - (aq) ↔ 2CrO 4 2- (aq) + H 2 O(l)
    • CH 3 CO 2 H(aq) + H 2 O(l) ↔ CH 3 CO 2 - (aq) + H 3 O + (aq)
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  • Chemical Equilibrium
    • Reactions not 100% complete
      • Products and Reactants exist together
    • A dynamic equilibrium
    • Position of equilibrium ???
    • Can the position of equilibrium be changed?
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  • Le Chatelier’s Principle
    • When an external change is made to a system in equilibrium, the system will respond to oppose the change
    • External Changes
    • Concentration
    • Pressure (gases)
    • Temperature
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Link to external video Link to external video
  • Concentration
    • 1. BiCl 3 (aq) + H 2 O(l) ↔ BiOCl(s) + 2HCl(aq)
    • 2. Cr 2 O 7 2- (aq) + 2OH - (aq) ↔ 2CrO 4 2- (aq) + H 2 O(l)
    • How does reaction 1 respond to addition of hydrochloric acid?
    • How does reaction 2 respond to addition of alkali?
    • How does reaction 2 respond to addition of acid?
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  • Pressure
    • N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)
    • CO(g) + 2H 2 (g) ↔ CH 3 OH(g)
    • 2NO 2 (g) ↔ 2NO(g) + O 2 (g)
    • PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g)
    • H 2 (g) + I 2 (g) ↔ 2HI(g)
    • CO(g) + H 2 O(g) ↔ CO 2 (g) + H 2 (g)
    • How do the above equilibria respond to:
      • An increase in pressure
      • A decrease in pressure
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  • Temperature
    • N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)  r H = -92.2 kJ mol -1
    • H 2 (g) + I 2 (g) ↔ 2HI(g)  r H = -9.4 kJ mol -1
    • CO(g) + H 2 O(g) ↔ CO 2 (g) + H 2 (g)  r H = -41.2 kJ mol -1
    • PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g)  r H = 87.9 kJ mol -1
    • How do the above respond to an
    • Increase in temperature
    • Decrease in temperature
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  • Equilibrium constants a measure of equilibrium position
    • aA + bB ↔cC + dD
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License BiCl 3 (aq) + H 2 O(l) ↔ BiOCl(s) + 2HCl(aq) Write the expressions for K c for the reactions given in previous slides
  • Calculating Equilibrium Constants
    • HNO 2 (aq) ↔ H + (aq) + NO 2 - (aq)
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License The table shows the equilibrium molar concentrations for three solutions of nitrous acid in water at 25 o C Calculate the equilibrium constant for this reaction at 25 o C Solution [HNO 2(aq) ] mol litre -1 [H + (aq) ] mol litre -1 [NO 2 - (aq)] mol litre -1 A 0.090 6.2 x 10 -3 6.2 x 10 -3 B 0.20 9.3 x 10 -3 9.3 x 10 -3 C 0.30 11.4 x 10 -3 11.4 x 10 -3
  • This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Units of K c Solution A Now try for solutions B and C
  • Acids and Bases This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  • Outline
    • Definitions
    • Weak Acids
    • Dissociation Constants
    • Weak Bases
    • Drugs
    • pH
    • Buffers
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  • Acids and Bases
    • Several definitions available - most common is Bronsted and Lowry
    • Acid is a proton donor
      • HCl is able to transfer H +
    • Base is a proton acceptor
      • NH 3 is able to accept H + and become NH 4 +
    • Aqueous solutions
    • Proton species is H 3 O + (hydroxonium ion)
      • HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq)
      • HCl(aq)  H + (aq) + Cl - (aq)
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  • Strong Acids
    • Strong acids are fully dissociated
    • HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq)
    • all dissolved HCl molecules are ionised
    • 1 mol dm -3 HCl (aq) there are:
      • Approx 1 mol dm -3 H 3 O + (aq)
      • Approx 1 mol dm -3 Cl - (aq)
    • DO NOT confuse ‘strong’ and ‘concentrated’
    • 1 x 10 -4 mol dm -3 HCl (aq) is a dilute solution of a strong acid
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  • Other strong acids
    • HNO 3 (nitric)
    • H 2 SO 4 (sulfuric)
    • HClO 4 (perchloric)
    • Write equations showing the dissociation of the above acids
    • Which are monoprotic?
    • Are any diprotic?
    • Chemical equilibrium – K very large
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  • Weak Acids
    • Acids that dissociate in a reversible reaction (e.g. CH 3 COOH; ethanoic (acetic) acid)
    • CH 3 COOH (aq) + H 2 O (l) ↔ H 3 O + (aq) + CH 3 COO - (aq)
    • Solution of CH 3 COOH (aq) contains:
      • CH 3 COOH (aq)
      • H 3 O + (aq)
      • CH 3 COO - (aq)
    • CH 3 COOH is partially dissociated
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  • How weak is a weak acid?
    • 0.1 mol dm -3 HCl is dissociated 91.4%
    • [H 3 O+] = 0.091 mol dm -3 pH=1.04
    • 0.1 mol dm -3 CH 3 COOH is dissociated 1.34%
    • [H 3 O + ] = 0.0013 mol dm -3 pH=2.87
    • Extent given by K
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  • Weak Acids
    • HA (aq) + H 2 O (l) ↔ H 3 O + (aq) + A - (aq)
    • HA Bronsted acid
    • H 2 O Bronsted base
    • H 3 O + Bronsted acid
    • A - Bronsted base
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  • Acid dissociation constant (K a )
    • The higher the K a value:
      • greater degree of ionisation
      • stronger the acid
      • Data tables
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  • K a Values
    • HCO 2 H 1.8 x 10 -4 mol dm -3
    • CH 3 CO 2 H 1.7 x 10 -5 mol dm -3
    • Are these weak or strong acids?
    • Which is the stronger acid?
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License HCO 2 H 3.75 CH 3 CO 2 H 4.77
  • pK a values (data tables) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Acid pK a Conjugate base H 3 PO 4 2.12 H 2 PO 4 - HNO 2 3.34 NO 2 - H 2 CO 3 6.37 HCO 3 - HCN 9.31 CN - HCO 3 - 10.25 CO 3 2-
  • pK a Values
    • Controlling the ionisation of weak acids
    • pH = pKa then [HA] = [A-]
    • pH > pKa then [A-] > [HA]
    • pH < pKa then [HA] > [A-]
    • CH 3 COOH (aq) + H 2 O (l) ↔ H 3 O + (aq) + CH 3 COO - (aq)
    • CH 3 COOH: CH 3 COO - at pH = 4.77 ?
    • CH 3 COOH: CH 3 COO - at pH = 3 ?
    • CH 3 COOH: CH 3 COO - at pH = 7 ?
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  • Henderson-Hasselbach This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License For weak acids Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?
  • Weak Bases
    • B (aq) + H 2 O (l)  BH + (aq) + OH - (aq)
    • CH 3 NH 2(aq) +H 2 O (l) ↔ CH 3 NH 3 + (aq) + OH - (aq)
    • pK a = 10.66 (of conjugate acid) [B]=[BH + ]
    • pH = 10.66
    • pH =8 what happens to CH 3 NH 3 + (aq) : CH 3 NH 2(aq)
    • pH =13 ?
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  • Henderson-Hasselbach This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License For weak bases Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?
  • Acidic drugs
    • 2-[4-(2-methylpropyl)phenyl]propanoic acid
    This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License How does this molecule ionise? pK a =4.5 pH =3 (stomach pH)? pH=6 (intestine)? ibuprofen
  • Basic drugs
    • amphetamine ( C 6 H 5 CH 2 CH(NH 2 )CH 3 )
    • Write an equation for the reaction of amphetamine with water.
    • The pK a of the conjugate acid is 9.8. What will happen to the ratio of ionised to unionised amphetamine at:
      • pH 7
      • pH 12
    • Why might this be important?
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  • Water
    • Can dissociate:
    • H 2 O (l) ↔ H + (aq) + OH - (aq)
    • 2H 2 O (l) ↔ H 3 O + (aq) + OH - (aq)
    • H 2 O is amphoteric
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  • Water
    • K w = [H 3 O + ][OH - ]= 1 x 10 -14 mol 2 dm -6
    • K w the ionic product of water
    • In pure water what is [H 3 O + ] and [OH - ] ?
    • K w is a very small constant
      • water is only very partially ionised
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  • pH
    • pH is defined as:
    • pH = -log 10 [H 3 O + ]
    • pH is a measure of the H 3 O + concentration in solution and can vary from 1 to 14
    • pH=7 – neutral [H 3 O + ] = [OH - ]
          • = 1 x 10 -7 mol dm -3 at 25 o C
    • pH<7 – acidic [H 3 O + ] >[OH - ]
    • pH>7 - alkaline/basic [H 3 O + ] <[OH - ]
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  • pH-examples
    • 0.1M HNO 3
    • 0.1M CH 3 COOH
    • What is the pH?
    • pH is dependent on the ionisation of the acid
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  • pH-examples
    • What about alkaline solutions?
    • E.g. 0.1M NaOH solution
    • Will also depend on degree of ionisation
    • use equation: [H + ] x [OH - ] = 10 -14
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  • Buffers This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License
  • Buffers
    • A buffer solution resists pH changes on addition of small amounts of acid or base (alkali) to a system .
    • Very important
      • e.g. blood has a pH of 7.4. If it varies by ± 0.4, death can occur
    • Buffer solutions rely upon the effects of a weak acid or base and the salt of that acid or base
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  • Buffers
    • Ethanoic acid (a weak acid) and sodium ethanoate (salt)
      • CH 3 COOH  CH 3 COO - + H + (1)
      • CH 3 COONa  CH 3 COO - + Na + (2)
    • (1)-partially ionised
    • (2)-fully ionised
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  • Buffers This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Henderson-Hasselbach equation Acidic buffers
  • Making a buffer solution
    • Choose a weak acid with a pK a close to the required pH of the buffer.
    • Choose an appropriate salt of the weak acid
    • Determine [salt]/[acid] ratio needed to give correct pH
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  • An acidic buffer: Ethanoic acid and sodium ethanoate This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License What is the [salt] if the acid is 0.1 mol dm -3 to give buffer solutions of pH = 5 pH = 4
      • What would be the pH of an ethanoate buffer with equal acid and sodium ethanoate concentrations?
  • An alkaline buffer: ammonia solution and ammonium chloride This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Note the base/salt ratio What is the pH of a buffer with base:salt ratio = 1? Calculate the base:salt ratios for pH 8.5 and pH 10.5
  • This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Acknowledgements
    • JISC
    • HEA
    • Centre for Educational Research and Development
    • School of natural and applied sciences
    • School of Journalism
    • SirenFM
    • http:// tango.freedesktop.org