CHEM2770 Lecture #2 - Water PDF

Lecture #2 Water  Water Chapter 2 - Water Water exists as a hydrogen bonded Chapter network 2 - Water Water – Ice has an average of 4exists as a...

Lecture #2 Water  Water Chapter 2 - Water Water exists as a hydrogen bonded Chapter network 2 - Water Water – Ice has an average of 4exists as abonds hydrogen H-bonded per water molecule network with an average of O Water exists – Liquid water has4 an as a H-bonded H-bonds average per molecule of 3.4 hydrogeninbonds H ice per O H waternetwork with an average of moleculeand o 3.4 in liquid. 4 H-bonds per molecule in ice H 104.5 H and 3.4 in liquid. o 104.5  Hydrogen Bonding H-bond: Hydrogen An electrostatic bonds attraction are electrostatic betweenbetween attraction polarized molecules polarized molecules containing containing O-H, O-H, N-H, N-H, or F-H or F-H. Hydrogen bonds H-bonds are strongest are strongest when when linear i.e.linear the two heavy atoms and the shared – Two H are heavy in aand atoms line.the shared hydrogen are in a line Hydrogen bonding H-bonding is a weak is a Weak interaction Interaction compared compared to covalentto covalent bonding - bonding 8-21 (8-21kJ/mol) kJ / mol. C 712 kJ / mol O 20 kJ / mol H 389 kJ / mol N  9/8/2014 Electrostatic Interactions 9/8/2014 Electrostatic interactions are attraction between oppositely charged Electrostatic Interactions: Attraction between oppositely charged ions or ionsrepulsion or repulsionbetween like charged between similarly chargedions ions - up to 200 kJ / mol. Up to 200kJ/mol –Electrostatic Interactions: Attraction between oppositely charged ions or repulsion between similarly charged ions - up to 200 kJ / mol. InSome –Some cells important important important hydrogen H-bond H-bond bond donors donors donors andand and acceptors acceptors acceptors in in cells: include: cells: O H C O N H C OH NH OH H N P O  Breaking H-bonds requires the addition of enthalpy. Breaking H-bonds requires the addition of enthalpy. Breaking H-bonds requires th For ice ∆HMelting For ice ∆H Melting = +6Hydrogen Bonds = +6kJ/mol kJ/mol For ice ∆HMelting = +6 kJ/m Breaking H-bonds requires the addition of enthalpy. Breaking hydrogen So why So whywater does does water sobonds meltmelt easilyrequires soeasily atat2525oC?oC? the addition of enthalpy For ice: So why does water melt so ea For ice ∆HMelting = +6 kJ/mol Because the liquid Because So why does iswater more the liquid disordered is more meltdisordered than than so easily the the atsolid 25 osolid and C? and Because the liquid is more di So why does water melt so easily at 25 oC? T∆Smelt > ∆H melt > ∆H. melt. melt > ∆Hmelt. – T∆S water has more disorder than solid water and T∆S Liquid melt atAt2525oC: In fact,–Because oC: theo T∆S liquid is melt = 6.6disordered more kJ/mol than the solid In andfact, at 25oC: T∆Smelt = In fact, at 25 C: T∆Smelt = 6.6 kJ/mol ∆HmeltT∆S = 6.0 kJ/mol > ∆Hmelt and. ∆Hmelt = 6.0 kJ/mol and ∆Hmelt = 6.0 kJ/mol and melt ∆Gmelt = ∆Hmelt – T∆Smelt = ∆Gmelt = ∆Hmelt – T∆Smelt = – 0.6 kJ/mol In∆G meltat= ∆H fact, oC:– T∆S 25melt T∆S = – 0.6 meltmelt kJ/mol = 6.6 kJ/mol ∆Hmelt So melting is an= entropy-driven 6.0 kJ/mol and processes. So melting is an entropy-dri – Melting is an entropy driven process ∆GSomelt = ∆H melting is anmelt entropy-driven – T∆Smeltprocesses. = – 0.6 kJ/mol  9/8/2014 Water 9/8/2014 Biomolecules interact with water by: Biomolecules interact with water by: 1. Biomolecules Hydrogen bonding interact with water by: H OO H 1. 1.H-bonding. H-bonding. O OH H H H SerSer 1. 2. Electrostatic Electrostatic interactions: interactions 2. Electrostatic i. i.WhenWhen interactions: NaCl NaCl dissolves dissolves in H2in O, water, enthalphy enthalpy is to is required required to+ break Na+ and Cl- break Na Cl- ionicbonds. bonds.+DH+∆H i. When NaCl dissolves in H2O, enthalpy is required to break Na+ Cl- ionic bonds. +∆H  Water ii. Enthalpy is also required toisdisrupt ii. Enthalpy also requiredhydrogen bonding to disrupt H–bonding of water of H2O. +∆H +DH iii. Enthalpy is released when new water–ion interactions form. ii. Enthalpy is releasedThis whenis callednew water-ion – “solvation”. –∆H interactions form –DH – Called solvation H H O H O H H H H O + O Cl - O Na H H H H O O H H H iv. Net enthalpy change iv. is Thesmall andchange net enthalpy slightly is smallpositive and slightly positive. v. Solid NaCl is highly ordered. NaCl in solution is highly iv. Solid NaCl is highly disordered. ordered.So NaClthe largein solution entropy increase is highly favours disordered. dissolution. So a large entropy increase favors dissolution  Water 9/8/20 ∆G = ∆H – T∆S ∆G = ∆H – T∆S T∆S >> ∆H and ∆G < 0 T∆S >> ∆H and ∆G < 0 3. Waals 3. van der van der Interactions: Waals Interactions: shortA short range,range veryweak very weak attractions 3. van der~Waals attraction Interactions: A short range very weak 4 kJ / mol. – ~4kJ/mol attraction ~ 4 kJ / mol. – Non-polar helium atoms form a liquid at 4K due to an induced dipole attraction Non-polar Non-polar HeHe atoms atomsform formaaliquid liquidatat4K 4Kdue duetoto ananinduced induced dipole dipole attraction. attraction. +- +- +- +- + - +- +- - Temporary Dipoles +  Water Non-polar hydrocarbons Non-polar hydrocarbons interact interact with each with each other by van by other der van der Waals interactions Waals interactions. CH3 CH3 CH2 CH2 CH3 CH3  Water What happens when a non-polar hydrocarbon is What happens when a non-polar hydrocarbon dissolves in water dissolved in water? 1. Hydrocarbon vdW interactions are broken. +∆H 1. Hydrocarbon van der Waals interactions are broken +DH 2. Water H-bonds are broken. +∆H 1. Water hydrogen bonds are broken +DH H 3. New 1. New water hydrogen water bonds H–bonds are formed in anare organized H H O cage around theformed in an organized hydrocarbon. “cage” This optimizes van der H Waals interactions between the hydrocarbonThis and water, O H around the hydrocarbon. CH3 H O and optimizes the hydrogen bonding among the water optimizes the vdW interactions molecules -DH H CH2 between the hydrocarbon and water, H O CH and optimizes the H–bonding 3 O H among the water molecules. –∆H O H H H  Water The entropy of water is reduced, disfavoring the dissolution of hydrocarbons in water –DS – This is called the Hydrophobic Effect Amphipathic molecules contain both polar and non-polar groups – Examples include detergents, lipids, proteins and nucleic acids – Their lowest free energy states have hydrophobic groups clustered together away from water, raising the water entropy – They help organize detergent micelles, membranes, proteins and DNA  Water A detergent A detergent:sodium dodecylsulphate Sodium dodecylsulphate: 9/8/2014 O Na+ -O S O CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 A detergent: Sodium dodecylsulphate: O O Na+ -O S O CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 It forms a micelleO in which the hydrocarbons interact with each other It forms via van derIta Waals micelle intowhich form forms a micelle theahydrocarbons hydrophobic in which interact core the hydrocarbons with and interact each the with eachhydrophilic otherassociate groups viaother vdWviato vdW form withtoa water form a hydrophobic core and the hydrophobic core and the hydrophilic groups associate with water. hydrophilic groups associate with water.  Water Other Other interactions cancan interactions occur with occur withthe the ionized formof ionized forms ofwater. water - H2O OH + H+ - Keq = [H+] [OH- ] = 1.8 x 10 16 M @ 25oC [H2O] [H2O] = 55.5 M (moles / L) So (Keq x 55.5) = [H+][OH- ] = 10-14 M2 = Kw Knowing two, you can solve for the third. A convenient way to express [H+] is pH. pH = – log10[H+]  Water In pure water: [H+] = 10-7 Molar So –log10[10-7] = – (–7) = 7 = pH pOH = – log10[OH-] In pure water – [OH-] = 10-7 Molar –log10[10-7] = – (–7) = 7 = pOH When pH = pOH we say water is “neutral”  pH can range from 0 to 14. pH pH can range from 0 to 14. Below pH 7 water is “acidic” – pH can range from 0-14 Below pHgastric Below pH=7 7waterjuice isispHis water 1. “acidic” – acidic gastric juice – Gastric juice is pH 1. is pH=1 Above pH=7 it pH Above is basic 7 it is “basic” - Above pH 7 – Egg white it is “basic” - pH=8 egg white is pH 8. egg white is pH 8. Strong acids: Strong Strong Acids – HCl is Acids a strong acid HClHCl isisa strong acid. acid. a strong - HCl HCl HH++ ++Cl- Cl 0.1 M 0.1 M + 0.1 M 0.1 M The acid dissociates 0.1 Mcompletely + 0.1 M and the pHacid – The The ~ 1. acid dissociates dissociates completelyand completely and the pH =1 the pH ~ 1.  Strong Bases NaOH is a strong base. - NaOH Acids and Bases ++ Na OH 1M 1M + 1M StrongStrong basesBases NaOH is a strong base. – NaOH is a strong base Complete dissociation NaOH gives pOH Na ~ 0+ pH ~ 14.- + OH Remember 100 = 1 1 M 1M + 1M – Complete dissociation gives pOH~0 pH~14 Complete dissociation gives pOH ~ 0 pH ~ 14. – Most Mostbiological acidsacids biological and bases andare weak.arei.e.weak bases they undergo 0 Remember incomplete 10 = 1incomplete dissociation Theydissociation. undergo - + COOare +weak. COOH acids and bases Most biological H i.e. they undergo incomplete dissociation. Weak Acid Conjugate Base - COOH COO + H+ A weak Aacid weakis aacid donoris aofdonor protons.of protons Weak Acid Conjugate Base A weak acid is a donor of protons.  Weak Acids- and Bases NH2 + H2O NH3+ + OH + - NH2 + H2O NH -+ OH NH3+ +Conjugate 3 NH2 + base Weak H2O OH Acid Weak base Conjugate Acid Weak base Conjugate Acid A weak base is an acceptor of protons. weak base isA an acceptor weak baseofis protons. an acceptor of protons A weak base is an acceptor of protons. Biochemists choose to express all weak acids and bases as weak acids Biochemists ochemists choosechoose to express to express all weak all weak acidsacids and and basesbases as “weak as “weak therefore: acids”, ds”, so:so:choose to express all weak acids and bases as “weak Biochemists + + ++ + acids”, so: NH NH 3 3 NH NH+ 2 2 H H Weak Weak NH Acid + Acid Conjugate NH + BaseBase Conjugate + H 3 2 general, In general, In Weak Acid general: Conjugate Base In general, - - + HA A + H + HA A + H - HA A + H+  Weak Acids and Bases Keq = [H+] [A- ] = Ka - HA A + H+ [HA] Ka is the acid dissociation constant. pKa = – log10[Ka] The weakest acids have the largest pKa’s. The pKa of acetic acid is 4.76, the pKa of ammonia is 9.25.  Weak Acids and Bases Often, biochemical reactions release H+ or OH-. How do cells deal with this? With Buffers. Often biochemical reactions release H+ or OH- - Often, biochemical reactions release + or OH. How do cells Hdifferent Titration – The cellCurves indicate deals with these the pH values situations using of buffers mixtures of deal with this? With Buffers. a weak acid and its conjugate base. The mixtures are produced by – adding Titrationdifferent amounts curves indicate theofpHa values strongofacid or strong different base of mixtures to weak the acid and Titration its conjugateCurves indicate base. The mixturesthearepH valuesby produced ofadding different mixtures different of of amounts mixture. a weak acid and its conjugate base. The mixtures are produced a strong acid or a strong base to the mixture by adding different amounts of a strong acid or strong base to the mixture. The pH of a 0.1 M solution of CH3COOH The is about 1.8. pH of a 0.1 M solution of CH3COOH is about 1.8. Addition of 10 mL of 0.1 M Addition NaOH to 100of mL10ofmL 0.1of M0.1 M NaOH to 100 mL of 0.1 M CH3COOH raises the pH to 3.7. CH3COOH raises the pH to 3.7. What happened? What happened?  Weak Acids and Bases 10 mL of 0.1 M NaOH contains 0.001 moles = 1 millimole of OH-. 0.1moles x 0.1moles = x = 10ml = 0.001moles 1000ml 10ml 1000ml 100 mL of 0.1 M CH3COOH contains 10 millimoles of acid. So we added 1 millimole or 0.1 equivalent of OH- = 10% of the acid present. Some of the added OH- reacted with H+ to yield H2O. The Law of Mass Action (AKA LeChatelier’s Principle) then caused the HA to ionize to release more protons. This continued until all the added OH- was neutralized. - HA A + H+  Weak Acids and Bases So [HA] has decreased, [A-] increased, [H+] decreased, [OH-] increased, and pH increased. The added OH- (a strong base) will neutralize an equal # of moles of HA (a weak acid). Similarly, a strong acid will neutralize an equal # of moles of conjugate base (A-). When 50 mL of NaOH have been added, 5 millimoles of OH- have been added, neutralizing half of the HA. The acid is half neutralized and so half dissociated [HA] = [A-] i.e. [Weak Acid] = [Conjugate Base]  Weak Acids and Bases Remember, [H + ][A− ] K a[HA] Ka = [H + ] = [HA] [A− ] So when [HA] = [A-] , [H+] = Ka and pH = pKa Near the end of the curve, when there is no more HA to ionize, the pH will rise sharply. The same graph could have been produced by adding HCl (a strong acid) to a solution of sodium acetate, the conjugate base of acetic acid (a weak base). Features of titration curves. 1. At the ends of the curves small additions of acid and base result in large changes in pH.  Weak Acids and Bases 2. In the middle, small additions of strong acid or base cause small changes in pH. The region where pH = pKa is called the buffering region. A Buffer is a mixture of WA and CB that resists changes in pH when small additions of strong acid or base are added. How does it work? Near the pKa [WA] ~ [CB]. The WA can neutralize added OH-, and the CB can neutralize added H+. Cells require buffers because high concentrations of H+ and OH - can break covalent bonds.  Weak Acids and Bases The main buffer system found in cells is : - - - H2PO4 + OH HPO42 + H2O 2- + + - HPO4 H H 2PO4 The pKa of H2PO4- is 7.2.  Weak Acids and Bases K a[HA] We can re-write [H + ] = − in terms of pH and pKa [A ] [HA] pH = pK a − log10 { − } [A ] Or [ A− ] pH = pK a + log10 { } [HA] Using this Henderson-Hasselbach Equation we can calculate the pH of a WA-CB pair if we know their ratio and the pKa. 1 [CB] 10 WA-CB pairs buffer well as long as: < < 10 [WA] 1  Weak Acids and Bases pH = pK a + log10 { } = pK a −1.0 pH = pK a + log10 { } = pK a +1.0 For acetic acid the range is: 3.76 to 5.76. Example Calculation: What is the pH of a 20 mL solution of 0.1 M Tris base (RNH2) after addition of 10 mL of 0.1 M HCl? The pKa = 8.1 First we calculate that we have 2 mmoles of Tris base and 1 mmole of HCl.  Weak Acids and Bases The reaction will be: The reaction will be: 2RNH2 + 1HCl + + - 1 RNH2 1 RNH3 + 1Cl This means that 1mmole of strong acid has neutralized 1mmole of The above means that 1 mmole of strong acid has neutralized 1 conjugate base mmole of conjugate base giving 1 mmole of WA and 1 mmole of – This produces 1mmole of weak acid and 1mmole of conjugate base CB. 1mmol / 30ml – HH: pH= 8.1 + Log 1mmole/30mL From HH: pH = 8.1+ Log{ 1mmole/30mL } 1mmol / 30ml – pH= 8.1 + 0 = 8.1 So pH = 8.1 + 0 = 8.1