Acids, Bases, pH & Buffers Lecture Notes PDF

Acids & Bases, pH and Buffers Dr Kassim Adebambo [email protected] Adapted from Dr.Mullane’s slides Blood pH levels © Science Media Group. 4LMS0004,  Learning...

Acids & Bases, pH and Buffers Dr Kassim Adebambo [email protected] Adapted from Dr.Mullane’s slides Blood pH levels © Science Media Group. 4LMS0004,  Learning Outcomes By the end of these two lectures and a period of independent study, students should be able to:  Explain the concept of acids and bases, pH and pOH, pH scale, strong and weak acids, importance in Biology  Understand auto-ionisation of water, Kw and pKw  Describe acid-base equilibrium, conjugate acid/base pairs, Ka, Kb, pKa, pKb  Perform pH calculations  Explain Buffer systems, and factors affecting choice of buffer,the Henderson-Hasselbalch 2 equation, and perform buffer calculations  Suggested Reading Chemistry for the Biosciences, J. Crowe & T. Bradshaw (3rd Ed.)(available as e-book in online library) Chapter 17 “Acids, bases and buffer solutions” “Catch up Chemistry” also v useful 3  Right now, in our bodies… Biochemical reactions are especially sensitive to pH. Saliva: pH of the reaction pH 6.5 medium must be controlled. Such control is provided by buffer solutions, which are Sweat: Stomach: solutions that maintain pH 5 pH 1-3 a particular pH. Two important biological buffer systems are the dihydrogen phosphate Blood: system and the pH 7.4 Intestine: carbonic acid system pH 6-7.5 Most drugs and biological molecules Urine: pH contain groups of atoms 5-8 that may be charged or neutral depending on 4 pH,  Acids and bases Electrolytes – on addition to water they produce ions – cations +; anions - The Brönsted-Lowry definition: An ACID is a PROTON DONOR HA H+ + A- A BASE is a PROTON ACCEPTOR H+ + B BH+ 5  Hydrogen Ions H2O H+ + OH- H+ cannot exist on its own in solution Solvated by another water molecule to give the hydronium ion, H3O+ H2O + H+ ⇌ H3O+ Wherever H+ and H3O+ in aqueous solution are mentioned, they mean the same thing 2H2O H3O+ + OH- Autoionisation : A reaction between two identical neutral molecules to produce a cation and an anion 6  Equilibrium For any equilibrium, we can generate an expression that shows how far to the right the equilibrium lies A + B C + D Equilibrium constant [C] [D] Keq = [A] [B] Square brackets indicate [concentration] 7  The water equilibrium The dissociation of water is an equilibrium reaction H2O H+ + OH- This equilibrium lies a long way to the left only about 1 in 10 million molecules will be dissociated Equilibrium constant is defined as [H+][OH-] Keq = [H2O] Notes: a) Square brackets indicate [concentration] b) [H+]=[H3O+] (see previous slide) 8  The water equilibrium H2O H+ + OH-  This equilibrium lies a long way to the left  The amount of ionisation is very small  [H2O] is much, much bigger than [H+][OH-]  [H2O] can be treated as a constant:  We can write a simpler expression: Kw = [H+][OH-] Where Kw is the ionisation product or ionic product of water 9 For pure water at 25oC [H+] = [OH-] = 1 x 10-7 M Kw = [H+][OH-] = 1 x 10-14 (M2) In neutral conditions [H+] = [OH-] ALWAYS In acid conditions [H+] > [OH-] ! In alkaline conditions [H+] < [OH-] Kw is an equilibrium constant, hence …..if [H+] goes up, [OH-] must go down pKw = - log Kw = 14 10  How do we measure acidity? We use the pH scale This is a logarithmic scale pH is a convenient way of expressing the concentration of H+ ions pH = – log [H+] 11  The pH scale We can define the acidity/alkalinity of a solution in terms of the H+ concentration pH = – log [H+] This is a logarithmic scale For low concentrations of H+ (less than 1M), -log [H+] is a positive number The pH scale is normally in the range 0 to 14 For a neutral solution: [H+] = 10-7M therefore pH = 7 For an acidic solution: [H+] > 10-7M and pH < 7 For an alkaline solution: [H+] < 10-7M and pH > 7 12  The pH scale Acidic Neutral Alkaline 13  A reminder about logarithms log = logarithm in base 10 log x = y where 10y = x for example log 1 = 0 (because 100 = 1) log 100 = 2 (because 102 = 100) log 0.001 = -3 (because 10-3 = 0.001) Get familiar with your calculator!! 14  A reminder about logarithms… Logarithms are particularly useful for ranges of very large or very small numbers. Using your calculator, Calculate the following: log 5000000 = log 0.5 = log 0.000002 = Log values can be converted back to integers by taking the inverse log (antilog) 10y = x 15  A reminder about logarithms Answers: log 5000000 = log 5 x 106 = 6.7 log 0.5 = - 0.30 log 0.000002 = log 2 x 10-6 = - 5.7 16  Example calculation 1 An aqueous solution contains 0.02 M of hydrogen ions. What is the pH of this solution? pH = – log [H+] pH = - (log 0.02) pH = - (-1.7) pH = 1.7 17  Now it’s Your Turn! Question: Calculate the pH of a 0.1M solution of hydrochloride acid, HCl Thinking caps on?? 18  Example calculation 2 What is the pH of a 0.1M solution of hydrochloride acid? HCl  H+ + Cl- pH = - log[H+] pH = - log(0.1) pH = 1 19  Sample exam question… What is the pH of a 0.1M solution of hydrochloride acid? a) 0.1 b) 7.0 c) 1.0 d) 2.5 e) 1.5 20  Example calculation 3 An aqueous solution contains 0.01 M hydroxide ions. What is the pH of this solution? Kw = [H+][OH-] = 1 x 10-14 10-14 [H+] = [OH-] [OH-] = 0.01 M gives us [H+] = 10-12 M continued… 21  Example calculation 3 ….continued An aqueous solution contains 0.01 M hydroxide ions. What is the pH of this solution? pH = – log [H+] pH = - (log 10-12) pH = - (-12) pH = 12 22  Equilibrium, acids and bases An acid, HA, can dissociate to give H+ ions This can be considered as an equilibrium reaction We can also think about how this acid reacts with a molecule of water – donating H+ HA H+ + A- HA + H2O H3O+ + A- 23  Equilibrium, acids and bases In the same way, we can think about a base B, that can accept H+ We can also think how the base behaves in water, removing a proton to generate OH- ions H+ + B BH+ H2O + B BH+ + OH- 24  Strong/Weak Acids and Bases HA H+ + A- For a STRONG acid, the equilibrium lies to the right - the acid completely dissociates into ions, e.g. hydrochloric acid HCl H+ + Cl- For a WEAK acid, the molecules only partially dissociate into ions, e.g. acetic acid CH3COOH H+ + CH3COO- 25  Strong/Weak Acids and Bases For a STRONG base, the base dissociates completely to generate a strong proton acceptor, e.g sodium hydroxide NaOH Na+ + OH- For a WEAK base, the molecules only partially dissociate into ions, e.g. ammonia NH3 + H2O NH4+ + OH- 26  Strong/Weak Acids and Bases BEWARE! Strong does not equal concentrated Weak does not equal dilute A 0.00001M solution of HCl is still a strong acid (but dilute) a 10M solution of ammonia is still a weak base (but concentrated) 27  weak strong HCl H+ + Cl- NaOH Na+ + OH- CH3COOH H+ + CH3COO- NH3 + H2O NH4+ + OH- 28  Conjugate Acids and Bases A weak acid dissociates to give [H+] and its CONJUGATE BASE The acetate ion is the conjugate base of acetic acid, as it can receive a proton to give the acid CH3COOH + H2O H3O+ + CH3COO- acid conjugate base 29  Conjugate Acids and Bases A weak acid dissociates to give [H+] and its CONJUGATE BASE The acetate ion is the conjugate base of acetic acid, as it can receive a proton to give the acid base acid CH3COOH + H2O H3O+ + CH3COO- acid conjugate In this reaction, water is acting as a base base 30  Conjugate Acids and Bases A weak base dissolves to give [OH-] and its CONJUGATE ACID The ammonium ion is the conjugate acid of ammonia, as it can lose a proton to give the base NH3 + H2O NH4+ + OH- base conjugate acid 31  Conjugate Acids and Bases A weak base dissolves to give [OH-] and its CONJUGATE ACID The ammonium ion is the conjugate acid of ammonia, as it can lose a proton to give the base acid base NH3 + H2O NH4+ + OH- base conjugate acid In this reaction water is acting as an acid 32  Acid equilibrium constant HA + H2O H3O+ + A- The equilibrium constant for dissociation of a weak acid HA in water is [H3O+][A-] Keq = [H2O][HA] 33  [H3O+][A-] Keq = [H2O][HA] But, as [H2O] can be treated as a constant, we can rewrite this as [H3O+][A-] Ka = ] [HA] Ka is known as the acidity constant 34  Similarly for a weak base H2O + B BH+ + OH- [BH+][OH-] Keq = [B][H2O] or [BH+][OH-] Kb = [B] Kb is known as the basicity constant 35  In the same way that we defined pH, we can define pKa = -logKa and pKb = -logKb these are useful for showing the strength of a weak acid or base 36  pKa = -logKa For example: CH3COOH H+ + CH3COO- Ka = 1.7 x 10-5 pKa = 4.75 The larger the value of Ka, the stronger the acid, and the smaller the value of pKa 37  pKb = -logKb For example: NH3 + H2O NH4+ + OH- Kb = 1.7 x 10-5 pKb = 4.75 The larger the value of Kb, the stronger the base, and the smaller the value of pKb 38  The relationship between Ka and Kb for an acid and its conjugate base acid HA + H2O H3O+ + A- conjugate base A- + H2O HA + OH- HA - acidity constant Ka, A- - basicity constant Kb KaKb = [H3O+][OH-] It can be shown that = Kw Hence: pKa + pKb = pKw pKa + pKb = 14 39 Hence we can use the pKa scale when talking about  Acid Strength Ka describes how much dissociation there is at equilibrium, i.e. how strong the acid is Ka values range over multiple powers of ten pKa is a way of showing acid strength in more convenient form Ka …10-1 10-5 10-10 10-15 10-20… pKa …1 5 10 15 20… acid strength Lower the pKa value = 40  Example Question 4 Which acid is the stronger acid? Ethanoic acid Ka = 1.8x10-5 Carbonic acid Ka = 4.2x10-7 41  Example calculation 4 Which is the strongest acid? pKa = -logKa Ethanoic acid Ka = 1.8x10-5 pKa = - Log 1.8x10-5 - (-4.75) = 4.75 Carbonic acid Ka = 4.2x10-7 pKa = - Log 4.2x10-7 - (-6.38) = 6.38 42 Remember…. Lower the pKa value = stronger a  Polyprotic (polybasic) acids HCl is a monoprotic acid - i.e. it can donate one proton Some acids can donate two or more protons - they are POLYPROTIC For example, sulphuric acid, H2SO4, dissociates to give H+ and the hydrogensulphate ion HSO4- The HSO4- ion can also dissociate to give H+ and the sulphate anion SO42- 43  H2SO4 is a strong acid (complete dissociation) H2SO4 H+ + HSO4- Ka is very large HSO4- is a weker acid and is at equilibrium, Ka = 1.2 x 10-2 hence pKa = 1.92 HSO4- H+ + SO42- The second deprotonation requires removal of a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral. 44  Other polyprotic acids Another polyprotic acid that is frequently used is phosphoric acid H3PO4 H+ + H2PO4- pKa 2.12 H2PO4- H+ + HPO42- pKa 7.21 HPO42- H+ + PO43- pKa 12.67 45  Buffers Normally, addition of a small amount of strong acid or base to an aqueous solution results in a large change in pH In biological systems in particular, it is important to keep pH within a certain narrow range A BUFFER solution is a solution which resists change in pH when acid or base is added 46 47  An example of a buffer Acetate buffer, containing acetic acid and sodium acetate in equal amounts CH3COOH + H2O H3O+ + CH3COO- Na[CH3COO] Na+ + CH3COO- The acid is mostly undissociated The salt (sodium acetate) produces sodium ions and acetate ions 48  How buffer solutions work CH3COOH + H2O H3O+ + CH3COO- The concentration of acetate ions from sodium acetate is much higher than acetate ions from the acid [CH3COOH] = initial concentration of acid [CH3COO-] = initial concentration of salt 49  How buffer solutions work CH3COOH + H2O H3O+ + CH3COO- 50  The buffering effect CH3COOH + H2O H3O+ + CH3COO- If we add some strong acid, the H+ ions will react with the reserve of acetate ions to form more acetic acid molecules The excess H+ ions are removed The effect of adding acid has been buffered 51  The buffering effect CH3COOH + H2O H3O+ + CH3COO- If we add strong base, the OH- ions will react with the H3O+ ions in solution More acid dissociates to balance the equilibrium The OH- ions are removed The effect of adding base has been buffered 52  pH of a buffer solution The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation [base] pH = pKa + log [acid] If the concentration of acid and conjugate base are equal, then pH = pKa The buffering effect works best when pH range is around pKa ± 1 53  Example calculation 5 A buffer solution contains sodium acetate and acetic acid, both at a concentration of 0.05 M What is the pH of the buffer? The pKa of aceteic acid is 4.75 [A-] pH = pKa + log [HA] NOTE: the base 10 logarithm of 1 is 0 Hence in this case pH = pKa 54  Choice of buffer solution We need our buffer solution to be at the desired pH for the experiment to effectively buffer the effect of any acid/base that is added during the experiment The amount of acid or base that can be added without a large change in pH is expressed as the BUFFER CAPACITY, b Buffer solutions work best in a pH range of around pKa ± 1 55  Choice of buffer solution Choose a system where the pKa of the acid is close to the desired pH e.g. For many biological experiments, a buffer of pH slightly above 7 is needed A “phosphate buffer” consists of dihydrogenphosphate as the______ (pKa 7.21) and hydrogenphosphate as the______ H2PO4- H+ + HPO42- 56  Example calculation 6 You are making up a buffer solution of dihydrogenphosphate (acid, pKa 7.21) and hydrogenphosphate (conjugate base). You require a pH of 7.35. What ratio of base to acid concentration do you need? Henderson-Hasselbalch equation [base] pH = pKa + log [acid] 57 (Further examples in workshop to follow…)  Acids and bases in biological systems One area where pKa and ionisation are important is in the absorption of medicines An oral dose of a medicine needs to be absorbed in the small intestine The first destination is the stomach, where the pH is very low - between 1 and 2 After the stomach is the small intestine, where the pH is around 6 Only neutral molecules can be efficiently absorbed into the bloodstream 58  Acids and bases in biological systems Most drug molecules have at least one acidic or basic group This means that they may be ionised, and the amount of ionisation depends on the surrounding pH The pH partition hypothesis states that unionised form of a drug can diffuse through the gastrointestinal membrane. A drug that is ionised (positively or negatively charged) will be very poorly absorbed through the gut wall 59  Acids and bases in biological systems If we know the pKa of a drug, we can calculate its how much is ionised at a particular pH From this we can work out if the drug will be absorbed in the stomach or the small intestine To do this we use the Henderson- Hasselbalch equation We must remember which is the ionised species! 60  Example Calculation 7 Aspirin is a weakly acidic drug, with pKa 3.25 Will aspirin be absorbed in the stomach? Assume stomach pH = 1 For an acid: [A-] pH = pKa + log [HA] Work on this at home: This will be covered in the workshop! 61  What you should understand Ionisation of water, Kw Concept of acids and bases, pH Acid-base equilibrium, conjugate acid/base pairs, Ka, Kb, pKa, pKb Neutralisation Buffer systems, the Henderson- Hasselbalch equation, calculations and factors affecting choice of buffer Why acids and bases are important in biology 62 Example MCQ question (see also the worksheet examples) What is the pH of a 0.0005M solution of nitric acid? a) 0.1 b) 0.76 c) 1.0 d) 7.6 e) 3.3 63