Chapter 2 - Water 9/8/2014. Water exists as a H-bonded network with an average of 4 H-bonds per molecule in ice and 3.4 in liquid. 104.

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1 Chapter 2 - Water Water exists as a -bonded network with an average of 4 -bonds per molecule in ice and 3.4 in liquid o -bond: An electrostatic attraction between polarized molecules containing -, N-, or F-. -bonds are strongest when linear i.e. the two heavy atoms and the shared are in a line. -bonding is a Weak Interaction compared to covalent bonding kj / mol. C N 712 kj / mol 20 kj / mol 389 kj / mol 1

2 Electrostatic Interactions: Attraction between oppositely charged ions or repulsion between similarly charged ions - up to 200 kj / mol. Some important -bond donors and acceptors in cells: C N N P Breaking -bonds requires the addition of enthalpy. For ice Melting = +6 kj/mol So why does water melt so easily at 25 o C? Because the liquid is more disordered than the solid and T S melt > melt. In fact, at 25 o C: T S melt = 6.6 kj/mol melt = 6.0 kj/mol and G melt = melt T S melt = 0.6 kj/mol So melting is an entropy-driven processes. 2

3 Biomolecules interact with water by: 1. -bonding. Ser 2. Electrostatic interactions: i. When NaCl dissolves in 2, enthalpy is required to break Na + Cl - ionic bonds. + ii. Enthalpy is also required to disrupt bonding of 2. + iii. Enthalpy is released when new water ion interactions form. This is called solvation. Na + - Cl iv. The net enthalpy change is small and slightly positive. v. Solid NaCl is highly ordered. NaCl in solution is highly disordered. So the large entropy increase favours dissolution. 3

4 G = T S T S >> and G < 0 3. van der Waals Interactions: A short range very weak attraction ~ 4 kj / mol. Non-polar e atoms form a liquid at 4K due to an induced dipole attraction Temporary Dipoles Non-polar hydrocarbons interact with each other by van der Waals interactions. C 3 C 2 C 3 C 3 C 2 C 3 4

5 What happens when a non-polar hydrocarbon dissolves in water? 1. ydrocarbon vdw interactions are broken Water -bonds are broken New water bonds are formed in an organized cage around the hydrocarbon. This optimizes the vdw interactions between the hydrocarbon and water, and optimizes the bonding among the water molecules. C 3 C 2 C 3 The entropy of water is reduced, disfavouring dissolution of hydrocarbons in water. S This is called the ydrophobic Effect. Amphipathic molecules contain both polar and non-polar groups. E.g. detergents, lipids, proteins, nucleic acids. Their lowest free energy states have hydrophobic groups clustered together away from the water, raising the water S. They help organize detergent micelles, membranes, proteins, and DNA. 5

6 A detergent: Sodium dodecylsulphate: Na + Ō S C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 3 It forms a micelle in which the hydrocarbons interact with each other via vdw to form a hydrophobic core and the hydrophilic groups associate with water. 6

7 ther interactions can occur with the ionized forms of water K eq = [ + ] [ - ] = 1.8 x o C [ 2 ] [ 2 ] = 55.5 M (moles / L) So (K eq x 55.5) = [ + ][ - ] = M 2 = Kw Knowing two, you can solve for the third. A convenient way to express [ + ] is p. p = log 10 [ + ] In pure water: [ + ] = 10-7 Molar So log 10 [10-7 ] = ( 7) = 7 = p p = log 10 [ - ] In pure water [ - ] = 10-7 Molar log 10 [10-7 ] = ( 7) = 7 = p When p = p we say water is neutral 7

8 p can range from 0 to 14. Below p 7 water is acidic gastric juice is p 1. Above p 7 it is basic - egg white is p 8. Strong Acids Cl is a strong acid. Cl + + Cl M 0.1 M M The acid dissociates completely and the p ~ 1. Strong Bases Na is a strong base. Na Na M 1 M + 1 M Complete dissociation gives p ~ 0 p ~ 14. Remember 10 0 = 1 Most biological acids and bases are weak. i.e. they undergo incomplete dissociation. C C Weak Acid Conjugate Base A weak acid is a donor of protons. 8

9 + N N Weak base Conjugate Acid A weak base is an acceptor of protons. Biochemists choose to express all weak acids and bases as weak acids, so: + N 3 N Weak Acid Conjugate Base In general, A A K eq = [ + ] [A - ] = K a [A] A A K a is the acid dissociation constant. pk a = log 10 [K a ] The weakest acids have the largest pk a s. The pk a of acetic acid is 4.76, the pk a of ammonia is

10 ften, biochemical reactions release + or -. ow do cells deal with this? With Buffers. Titration Curves indicate the p values of different mixtures of a weak acid and its conjugate base. The mixtures are produced by adding different amounts of a strong acid or strong base to the mixture. The p of a 0.1 M solution of C 3 C is about 1.8. Addition of 10 ml of 0.1 M Na to 100 ml of 0.1 M C 3 C raises the p to 3.7. What happened? 10 ml of 0.1 M Na contains moles = 1 millimole of moles = x 1000ml 10ml x=10ml 0.1moles = 0.001moles 1000ml 100 ml of 0.1 M C 3 C contains 10 millimoles of acid. So we added 1 millimole or 0.1 equivalent of - = 10% of the acid present. Some of the added - reacted with + to yield 2. The Law of Mass Action (AKA LeChatelier s Principle) then caused the A to ionize to release more protons. This continued until all the added - was neutralized. A A

11 So [A] has decreased, [A - ] increased, [ + ] decreased, [ - ] increased, and p increased. The added - (a strong base) will neutralize an equal # of moles of A (a weak acid). Similarly, a strong acid will neutralize an equal # of moles of conjugate base (A - ). When 50 ml of Na have been added, 5 millimoles of - have been added, neutralizing half of the A. The acid is half neutralized and so half dissociated [A] = [A - ] i.e. [Weak Acid] = [Conjugate Base] Remember, K a = [+ ][A ] [A] [ + ]= K a[a] [A ] So when [A] = [A - ], [ + ] = K a and p = pk a Near the end of the curve, when there is no more A to ionize, the p will rise sharply. The same graph could have been produced by adding Cl (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 p. 11

12 2. In the middle, small additions of strong acid or base cause small changes in p. The region where p = pk a is called the buffering region. A Buffer is a mixture of WA and CB that resists changes in p when small additions of strong acid or base are added. ow does it work? Near the pk a [WA] ~ [CB]. The WA can neutralize added -, and the CB can neutralize added +. Cells require buffers because high concentrations of + and - can break covalent bonds. The main buffer system found in cells is : 2 P P P P 4 - The pk a of 2 P 4 - is

13 We can re-write [ + ]= K a [A] [A ] in terms of p and pk a p =pk a log 10 { [A] [A ] } r p =pk a + log 10 { [A ] [A] } Using this enderson-asselbach Equation we can calculate the p of a WA-CB pair if we know their ratio and the pk a. WA-CB pairs buffer well as long as: 1 10 < [CB] [WA] < 10 1 p =pk a + log 10 { [1] [10] }=pk a 1.0 p =pk a + log 10 { [10] [1] }=pk a +1.0 For acetic acid the range is: 3.76 to Example Calculation: What is the p of a 20 ml solution of 0.1 M Tris base (RN 2 ) after addition of 10 ml of 0.1 M Cl? The pk a = 8.1 First we calculate that we have 2 mmoles of Tris base and 1 mmole of Cl. 13

14 The reaction will be: 2RN Cl 1RN 2 + 1RN 3 + 1Cl - The above means that 1 mmole of strong acid has neutralized 1 mmole of conjugate base giving 1 mmole of WA and 1 mmole of CB. From : So p = = 8.1 1mmol / 30ml p = 8.1+Log{ 1mmol / 30ml } 14