lecture 3.105

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 LAST TIME WE ENDED WHILE TALKING ABOUT WEAK FORCES, HYDROPHOBIC FORCES IN PARTICULAR. 4b. Hydrophobic forces = entropic forces. < 40 kJ/mol. Hexane and water won't mix. In general, hydrocarbons and water won't mix, because of hydrophobic forces. The root cause is the entropy (disorder) of water. Water becomes more ordered as it packs around a hydrocarbon. This causes a decrease in the system's entropy, and so is not favored energetically. What does water really look like? Flickering clusters What happens when you dissolve a non-polar molecule in water?

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 Ask yourself, how many water molecules are not free to go anywhere randomly in the two configurations? 12 x’s per molecule x 2 = 24 x’s vs. 16 x’s. If they hydrocarbons are in a solution with 100 water molecules, 76 water molecules are free on the left and 84 water molecules are free on the right. Break hydrogen bonds, but can't compensate as in salts by new interactions. In fact, water is constrained to line up in an ordered fashion around hydrocarbons, decreasing the system's entropy. Thus, increase in  G as hydrocarbons dissolve. Less entropy decrease by grouping the hydrocarbons together and keeping the water together. This is what causes water and oils not to mix. It isn't actually that hydrocarbons "like" to be together, but rather that water "likes" to exclude them. Back to the principle of the system seeking its lowest energy configuration. Molecules that act like these hydrocarbons in this example are said to be " hydrophobic ", i.e., 'afraid of water'. Molecules that are easily solvated by water are termed " hydrophilic ", i.e., 'water loving'. Hydrophobicity and hydrophilicity are properties that figure prominently in biochemistry. A working biochemist refers to these concepts regularly; you must understand it. Points to remember about forces holding molecules together: For covalent bonds, longer = weaker, mono<double<triple bonds. Non-covalent interactions are all considerably weaker than covalent bonds. van der Waals interactions are possible only at very short distances Why talk about non-covalent interactions? They add up to very large and important forces in macromolecules. That is, the sum of many non-covalent interactions is considerable.

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 4 For simplicity, we write the ionization reaction as: H 2 O  H + + OH - It has been determined that pure water at room temperature has [H + ] = 10e-7 M. By examination of the ionization reaction, we can see that [OH - ] = Define "p" function: - log 10 (something). For pure water at RT, p[H + ] = pH = 7.0. This is defined as neutral pH. Note that pH and other "p" functions are logarithmic. A change in pH of one unit = a ten-fold change in [H + ]. Even though the [H + ] is very small, 10e-7, it is critical for biochemistry. Changes in pH mediate many things in the cell, including protein structure, ATP synthesis, ligand/receptor interactions, etc. Biochemists need a clear understanding of pH. So, we are going to spend some time working with acid/base problems. Look at dissociation of water more closely (with k f /k r ): (We’re going to derive the ion product of water, which leads to the relation pOH = 14 - pH.) k f H 2 O  H + + OH - k r The k's are the rate constants. (The general way to write the rate equation of any reaction A k → B is: dA/dt = -k[A], and dB/dt = k[A]. This reads “ the change in concentration of A or B with time = ... ” ) The forward rate = k f [H 2 O], the reverse rate = k r [H + ][OH - ]. At equilibrium, the relationship is ... forward and reverse rates are equal. So, k f [H 2 O] = k r [H + ][OH - ] at equilibrium. 10e-7 M

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 5 Then k f /k r = [H + ][OH - ] / H 2 O = K d = the dissociation constant = the equilibrium constant This is true for all solutions of water, so must be true for pure water. K d = 10 -7 x 10 -7 / [H 2 O], where ( [H 2 O] = 1000 g/L / 18 g/mol = 55.6 mol/L) K d = 10 -14 / 55.6 = 1.8 -16 M Since [H 2 O] almost always constant in biochemistry, define a new constant K w = K d •[H 2 O] = [H + ][OH - ] = 10 -14 (@ RT) = the ion product of water This tells us that [H + ] and [OH - ] can change, but the product stays the same. So if we know [H + ], we know [OH - ]. pK w = -log(10 -14 ) = 14. Also, pK w = -log([H + ][OH - ) = -log( [H + ]) + -log([OH - ]) = pH + pOH Since pH + pOH = 14, if we know pH, we know pOH as 14 - pH. Acids and Bases: The acid is the H + donor, the base is the H + acceptor. A strong acid dissociates completely, as does a strong base . Take HCl as an example of a strong acid: HCl  H + + Cl - ; essentially complete reaction in water. (One direction only) So, what's the pH of a solution of 0.1 M HCl? [H + ] = protons from water dissociation (10 -7 M) + protons from HCl (0.1 M). Thus, pH = -log [H+] = -log(0.1 + 0.0000001) = - log (0.1000001) ≈ -log(0.1) = 1

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 7 CH 3 COOH  CH 3 COO - + H + K a = [H + ][CH 3 COO - ] / [CH 3 COOH] = 1.74 -5 M K a experimentally determined and can vary with T, [salt], ... Dissociation constants are often given as pK a 's, convenient. For acetic acid, pK a = -log(K a ) = -log(1.74e-5) = 4.76. Low pK a means more H + , stronger acid; high pK a means stronger base. Here’s how you calculate the pH of a solution of 0.1 M acetic acid. The same methodology applies to all weak acid and base problems. Think of it as adding 0.1 mol of acetic acid to 1 L of water, with starting and ending concentrations. ICE problem. CH 3 COOH  CH 3 COO - + H + starting concentration (Initial) 0.1 0 0 (Change) -x x x ending concentration (End or Equilib) 0.1 - x x x Put into equilibrium equation: 1.74 -5 = X  X / (0.1 - X), or X 2 + 1.74  10 -5 X -1.74  10 -6 = 0 Can simplify the equation considerably if assume that X is << starting concentration, i.e., 0.1, that is 0.1 - X  0.1. This assumption is ok (for me, personal decision) if the resulting error is below 5%. That is, if X / 0.1 < 0.05. Or more generally, if X / C T < 0.05, with C T being the total acid (conjugate acid + conjugate base). Must check this each time. Making this assumption (just try it), then, 1.74e-5 = X  X / 0.1, or X 2 = 0.1  1.74e-5 = 1.74e-6

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 8 Then, X = (1.74e-6) 1/2 = 1.32e-3 = [H + ] Check assumption: 1.32e-3/0.1 = 0.0132 = 1.32% error; assumption ok. So, pH = -log[H + ] = -log(1.32e-3) = 2.88. Very acidic. What would you have if the assumption of X << 0.1 did not hold? A quadratic equation of the form aX 2 + bX + c = 0, which can be solved by the quadratic solution Generally won't require this solution, but you must check it each time. We continue now to talk about the titration of a weak acid. Titration What happens if we add base, NaOH, to titrate the acid? First realize that since NaOH is a strong base, addition of NaOH is equivalent to the addition of OH - . CH 3 OOH + OH -  CH 3 COO - + H + + OH -  CH 3 COO - + H 2 O Or AH + OH-  A- Since adding OH - , pushing the chemical reaction to the right. How far to the right? Now there is less acid present, so the pH rises. pH titration curve ([NaOH] on abscissa, pH on ordinate):

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 10 This shows that the pH of a given conjugate acid/base pair is dependent on the ratio of these two species. This is the Henderson Hasselbalch equation , which has the general form: pH = pK a + log{conjugate base/conjugate acid} = pK a + log{[A - ]/[HA]} Let's derive the Henderson Hasselbalch equation again in a general way in five steps: HA  H + + A - K a = [H + ] [A - ] / [HA] [H + ] = K a  [HA]/[A - ] -log[H + ] = -log(K a ) -log([HA]/[A - ]) pH = pK a + log ([A - ]/[HA]) This is a mathematical description of the titration curve we saw earlier. For example, if add 0.02 moles of NaOH to the 1.0 L of 0.1 M acetic acid. CH 3 OOH + OH -  CH 3 COO - + H 2 O We know the starting concentration of CH 3 COOH is 0.1 M (- 1.32e-3 = 0.1). Since NaOH is a strong base, 0.02 moles of it will convert 0.02 moles of CH 3 COOH to 0.02 moles of CH 3 COO - , consuming the resultant H + to make water. So, the concentration of CH 3 COO - is 0.02 moles (per liter). pH = pK a + log[0.02 / (0.1-0.02)] = pK a + log (0.02/0.08) = 4.76 - 0.60 = 4.16 Another example. What if we added 0.05 moles NaOH?

Biological Sciences 105 Lecture 3, October 4, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 11 pH = pK a + log[0.05 / (0.1 - 0.05)] = pK a + log(0.05/0.05) = pK a + log(1) = pK a + 0 = 4.76 This demonstrates that when the concentration of acceptor and donor concentrations are the same, the pH is the pK a . Let’s say you add a number of moles of NaOH equivalent to ½ the number of moles of acid present, i.e., C T x volume. What is the pH? It is the pK a .