lecture 4.105

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Biological Sciences 105 Lecture 4, October 9, 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 TALKING ABOUT ICE PROBLEMS Another example. What if we added 0.05 moles NaOH? 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 . Then the pK a can be seen as the midpoint of the titration, i.e., the inflection of the sigmoidal curve. Look at the slope of the curve. It predicts that the amount of pH change resulting from the addition of a certain amount of acid or base is dependent on where you are on the curve. The ability to resist changes in pH upon the addition of acid or base is called "buffering", and buffers are compounds that do that. For acetic acid, where does it have the strongest buffering power? At the midpoint of the titration, i.e., at the pK a . This is true of all buffers. In fact, compounds have useful buffering powers at approx. 0.5 pH units on either side of the pK a 's. This is why the pK a of laboratory buffers are the parameters reported, as this is the important parameter required to choose an appropriate buffer for a particular application. Many biochemical reactions release or consume H + . It is important to buffer the cell or compartment so as not to allow the pH to enter non-physiological ranges. Buffering is critical.

Biological Sciences 105 Lecture 4, October 9, 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 Back to the titration: We saw where the numbers 2.88 and 4.76 come from. Now let’s look at the end of the titrations where we’ve added 0.1 M NaOH. That is, where we’ve added an equimolar amount of base to the acid. End of the titration . To do this problem, pretend that you are adding base to water. Then it’s much like the weak acid to water problem that we solved to get the number 2.88 at the beginning of the titration. First we’ll have to see what we mean by K b . To add a base to water, it is usually easiest to write it this way: A - + HOH  HA + OH - . The equilibrium constant is K eq = [HA][OH - ] / [A - ][HOH]. As with K a , lose the water term by multiplying the equilibrium constant by [HOH], to get the base dissociation constant, K b . − − = ] [ ] ][ [ A OH HA K b It’s as if we had written the base addition reaction as A -  HA + OH - .

Biological Sciences 105 Lecture 4, October 9, 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. 3 What’s the relationship between K a and K b ? K w = K a  K b = = 10 -14 Now you see that pK a + pK b = 14. True for all bases. That means that pK b for acetic acid is 14 – 4.76 = 9.24, and K b = 5.75  10 -10 . Back to the problem A - + H 2 O  HA + OH - Now set up as initial and final concentrations. A - + H 2 O  HA + OH - initial 0.1 [H 2 O] 0 0 change -x x x end 0.1-x [H 2 O] x x From the equilibrium we know ] )[ 1 . 0 ( 2 2 O H X X K eq − = But from our earlier considerations, this is also ) 1 . 0 ( ] [ 2 X X K K HOH b eq − = = Now, X X K b − = 1 . 0 2 , and this is simplified as we did for the original problem. First try the assumption that 0.1 – X  0.1. Then X 2 = 0.1K b = 5.75  10 -11

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Biological Sciences 105 Lecture 4, October 9, 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 X = 7.59  10 -6 = [OH - ] Now ask, is X / 0.1 < 0.05? Yes. pOH = 5.12 = 14 - pH pH = 8.88 Summarize some important relationships K a = [conj. base]  [H + ] / [conj. acid] K b = [conj. acid]  [OH - ] / [conj. base] K a  K b = 10E-14 pK a + pK b = 14 pH = pK a + log{[conj. base] / [conj. acid]} Another thing about the titration: Could have plotted on the abscissa the ratio of NaOH added to buffer present, i.e., ????? 𝑂𝐻 − ????? ????? ?? ?????? ???????

Biological Sciences 105 Lecture 4, October 9, 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 Mentioned earlier that acid/base pairs are used as buffers. Where? Around their respective pK a 's. So acetic acid is a reasonable buffer around pH 4 - 5. Most biochemistry takes place near neutral pH, so we need buffers with pK a 's between 6.5 and 8.5. Some rules of thumb What to do if you are asked to a dd acid or base to water, what’s the pH? Set up ICE problem, solve for X. What to do if you are asked to add H + or OH - to a buffer, determine the ratio of base to acid, titrations; use Henderson-Hasselbalch equation. Some sample problems. I will not do all these in class, but will leave them in the lecture notes for you to see if you wish: Example #1 : In a 1.0 L solution of 0.1 M Tris (pKa = 8.1) at pH 7.5, what is the concentration of protonated and unprotonated Tris? The solution comes from these steps: 1. Determine moles A - / moles HA from the H-H equation 2. Express A - as xHA 3. N T = A - + HA = x HA + HA = (1+x)HA 4. Plug in numbers 5. Check that it makes sense Use Henderson-Hasselbalch equation pH = pK a + log{A - /HA} 7.5 = 8.1 + log{A - /HA} log{A - /HA} = 7.5 - 8.1 = -0.6

Biological Sciences 105 Lecture 4, October 9, 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. 6 [A - ]/[HA] = 10^(-0.6) = 10 -0.6 = 0.25 [A - ] = 0.25[HA] N T = total number of moles of buffer (acidic and basic forms) = A - + HA 0.1 = 0.25[HA] + [HA] = 1.25[HA] [HA] = 0.1/1.25 = 0.08 M [A - ] = 0.1 - 0.08 = 0.02 M This answer makes sense since pH < pK a , and so should have more protonated than unprotonated buffer. Example #2 : How many moles of Tris base and HCl are required to make a 1.0 L solution of 0.1 M Tris-HCl at pH 7.8? Solution from these steps: 1. Solve for A - /HA using H-H equation. 2. Solve for A - = xHA 3. Replace A - with xHA in N T = A - + HA 4. Solve for numerical values of HA and A - 5. Need as many moles of acid as conjugate acid produced from conjugate base. 6. Check that the answer makes sense. For this example, pH = pKa + log(A - /HA) 7.8 = 8.1 + log(A - /HA) A - /HA = 10 -0.3 = 0.5 A - = 0.5  HA N T = 0.1 = A - + HA = 0.5  HA + HA = 1.5  HA

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Biological Sciences 105 Lecture 4, October 9, 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 HA + 0.1/1.5 = 0.067 So we need 0.067 moles of HCl to convert 0.067 of A - into HA. Example #3 : How many moles of Tris base and HCl are required to make a 1.0 L solution of 0.1 M Tris-HCl at pH 8.1? This can be solved by inspection. Since the pH is the pK a , we need to neutralize half of the Tris. So we need 0.1 moles of Tris base and 0.05 moles of HCl. Example #4: Add 0.005 moles H + per liter to 20 mM Tris at pH 7.8, what is the final pH? Solution from these steps: 1. What are the initial concentrations of HA and A - ? Get these from the Henderson-Hasselbalch equation. 2. Calculate [HA] i + 0.005 mol/L H + . That’s [HA ] f . 3. From that calculate [A - ] f . 4. Put the new ratios into the Henderson-Hasselbalch equation and calculate pH f . 5. Does this make sense? pH = pKa + log(A - /HA) 7.8 = 8.1 + log(A - /HA) A - /HA = 10 -0.3 = 0.5 A - = 0.5  HA N T = 0.02 = A - + HA = 0.5  HA + HA = 1.5  HA HA i + 0.02/1.5 = 0.0133 A - i = 0.02 – 0.0133 = 0.0067 HA f = HA i + 0.005 = 0.0183

Biological Sciences 105 Lecture 4, October 9, 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 A f = 0.0067 – 0.005 = 0.0017 pH = pK a + log { A - f /HA f } = 8.1 – 1.04 = 7.06 Example #5: It takes 16.7 mmol of HCl to bring a 1 L, 50 mM solution of TRIS base to pH 8.4. How much would it take to bring a 50 mM solution of TRIS base to pH 7.8? Can solve this by inspection. To get to pH 8.1, requires 25 mM HCl. To stop at pH 8.4 requires 25 – 16.7 = 8.3 mmol less than that required to get to the pKa. The answer asks you to go symmetrically below the pKa. So you’ll need 25 + 8.3 = 33.3 mmol. Polyprotic acids Physiological buffers maintain pH in the cell. One important physiological buffer is phosphoric acid/phosphate. The fully protonated form is It is different from other buffers we have discussed before because it has more than one proton that can dissociate. This is a polyprotic acid. The dissociation reaction is: H 3 PO 4  H 2 PO 4 - + H +  HPO 4 2- + H +  PO 4 3- + H + K a1 , K a2 , K a3 ; pK a1 = 2.1, pK a2 = 7.2, pK a3 = 12.3 As each H + comes off, the molecule becomes more negative. Therefore harder to remove next proton, making it a weaker acid. Look at titration of 0.1 M H 3 PO 4 in 1 L.

Biological Sciences 105 Lecture 4, October 9, 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. 9 The titration curve looks just like the individual titration of three monoprotic acids. Polyprotic acids Physiological buffers maintain pH in the cell. One important physiological buffer is phosphoric acid/phosphate. The fully protonated form is It is different from other buffers we have discussed before because it has more than one proton that can dissociate. This is a polyprotic acid. The dissociation reaction is: H 3 PO 4  H 2 PO 4 - + H +  HPO 4 2- + H +  PO 4 3- + H + K a1 , K a2 , K a3 ; pK a1 = 2.1, pK a2 = 7.2, pK a3 = 12.3 As each H + comes off, the molecule becomes more negative. Therefore harder to remove next proton, making it a weaker acid. Look at titration of 0.1 M H 3 PO 4 in 1 L.

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Biological Sciences 105 Lecture 4, October 9, 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 The titration curve looks just like the individual titration of three monoprotic acids. We can ask what are the concentrations of the various species at, say, pH 6.5? Essentially no H 3 PO 4 , because the pH investigated is too far from the pKa. 6.5 = 2.1 + log{[H 2 PO 4 - ] / [H 3 PO 4 ]} 4.4 = log{[H 2 PO 4 - ] / [H 3 PO 4 ]} Therefore, the ratio of base to acid about 10 4.4 , or 1:10,000. Insignificant. Same is true of PO 4 3- , its concentration is insignificant. That is, 6.5 = 12.3 = log([PO 4 3- ] / [HPO 4 2- ]) and the ratio is 10 -5.8 , or about 1/1,000,000. What of the two intermediates? 6.5 = 7.2 + log{[HPO 4 2- ]/[H 2 PO 4 - ]} [HPO 4 2- ]/[H 2 PO 4 - ] = 10 -0.7 = 0.2 = [A - ]/[HA] [A - ] = 0.2[HA] N T = 0.1 M = [A - ] + [HA] = 0.2[HA] + [HA] = 1.2[HA] [HA] = 0.1/1.2 = 0.083 = [H 2 PO 4 - ]

Biological Sciences 105 Lecture 4, October 9, 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 [A - ] = 0.1 - 0.083 = 0.017 = [HPO 4 2- ] Other species are negligible. We have just solved the question of the concentration of various species present at pH 6.5. Recall that the two outlying species were at concentrations too low to be considered significant. The internal two went into the Henderson-Hasselbalch equation to get their concentrations. What you must realize when working with polyprotic acids is which pK a to use in the calculation . It is the one between the conjugate acid/base pairs that you are considering. The problem we just solved asked what if we had set the pH to 6.5? Now let's ask what would happen if we just added one of the intermediates to water. You know how to solve the problem of calculating the pH when you add the fully protonated or fully deprotonated species to water. This is the acid or base problem we solved for acetic acid. What about a form that is not at the end? Formally, the question is, what is the pH if we add 0.1 moles of KH 2 PO 4 to water? This means that we are adding H 2 PO 4 - to water since the cation dissociates completely. The H + dissociation reaction is: H 2 PO 4 - + H 2 O  HPO 4 2- + H 3 O + ; pK a = 7.2 But the H + released can combine with the original: H 2 PO 4 - + H 3 O +  H 3 PO 4 + H 2 O; pK a = 2.1 This acts to reduce the effect of the initial H + extraction. The sum of the two reactions is: 2 H 2 PO 4 -  H 3 PO 4 + HPO 4 2- This is called a disproportionation reaction . When we add a disproportionating species to water, the resulting pH is simply ½(pK a1 + pK a2 ) = ½(7.2 + 2.1) = 4.65 HPO 4 - can act as an acid or a base. This makes it amphoteric . Since different charges are associated with different states of protonation, it is an amphoteric electrolyte , or ampholyte . What interesting biochemicals are ampholytes? Amino acids, proteins, nucleotides, nucleic acids.

Biological Sciences 105 Lecture 4, October 9, 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. 12 We are done with the chemistry, now to biochemistry. The first truly biochemical topic we'll discuss is proteins, their structure and function. In fact, we'll spend a lot of time on proteins. In fact, the study of proteins has dominated biochemistry for the past 80 years or so, and still does today. Why? Because almost nothing happens without the involvement of proteins at some point. Bacteria contain about 3,000 different proteins; humans, 30,000 or so. What do proteins do? 1. Enzymes. Enzymes are the catalysis for all metabolism. Almost nothing biochemical happens without being catalyzed (I can't think of anything). Most enzymes are proteins. 2. Transport proteins. Proteins carry molecules across cellular membranes, and throughout the organism. We'll focus on hemoglobin as an example later. 3. Nutrient and storage proteins. Proteins can be broken down for the chemical energy they contain. Seeds start out this way, ovalbumin (in egg white) and casein (in milk) have this function. 4. Contractile and motile proteins. Actin/myosin; kinesin; 5. Structural proteins. Some that you might have heard of are; collagen (tendons and cartilage), elastin (ligaments, 2d stretching), keratin (hair, nails, feathers), and more. 6. Defense proteins. Snake venoms, plant and bacterial toxins, antibodies. 7. Regulatory proteins. A major element in biochemical studies is how pathways and processes are regulated. Certainly a hot topic today. G proteins as one example. 8. Other exotic functions: In almost any biochemical activity, a protein has evolved to mediate it. If you wanted to create a new biochemical property, you'd probably start by designing a protein to do it. Antifreeze proteins, luciferase, aquorin, ...

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