# Bio 98 - Lecture 2 Acid-Base Equilibria, pH and Buffers.

## Presentation on theme: "Bio 98 - Lecture 2 Acid-Base Equilibria, pH and Buffers."— Presentation transcript:

http://en.wikipedia.org/wiki/Acid_Queen

Bio 98 - Lecture 2 Acid-Base Equilibria, pH and Buffers

pH profiles of enzymatic reactions UCI Bio199 Independent Research Pepsin Amylase

H 2 O H + + OH - Pure water is only slightly ionized H + ions (protons) do not persist free in solution, they are immediately hydrated to hydronium ions (H + + H 2 O H 3 O + ). Grotthuss proton wire

Achieving equilibrium K w = [H + ] [OH - ] = 10 -14 M 2 [H + ] = [OH - ] = 10 -7 M = 0.1  M [H + ] [OH - ] [H 2 O] Concentration of “water in water” ([H 2 O]) is 55.6 M [next slide], thus Pure water has equal quantities of H + and OH - ions, or, put differently, pure water has equal [H + ] and [OH - ]. Constant ion product! = 1.8 x 10 -16 M K eq = H 2 O H + + OH -

[H 2 O] Concentration is measured in moles per liter (mol/l) or simply M. 1 l = 1,000 ml of water has a mass of 1,000 g. 1 mole of water has a mass of 18 g (hydrogen 1 Da, oxygen 16 Da). Thus 1 liter of water (1,000 g) contains 1,000 g / 18 g moles of water. [H 2 O] = (1,000 g / 18 g) M = 55.6 M

pH = log [H + ] 1 = -log [H + ] pH = -log(10 -7 ) = -(-7) = 7 The pH scale In neutral/pure water [H + ] = [OH - ] = 10 -7 M, so Logarithm (base 10) refresher: if log 10 (x)=y then x=10 y Acid: proton donor Base: proton acceptor

pH = -log [H + ] HCl H + + Cl - HCl is a strong acid that completely dissociates in water. 1 M HCl will thus yield 1 M [H + ] and the pH will be pH = -log [H + ] = -log(1) = 0 NaOH is a strong base that completely dissociates in water. 1 M NaOH will thus yield 1 M [OH - ]. Since [H + ] [OH - ] = 10 -14 M and must remain constant [H + ] = 10 -14 M and the pH will be pH = -log [H + ] = -log(10 -14 ) = 14 Life is compatible only in a narrow pH range around pH 7. Strong acids and bases

Dissociation of a weak acid or weak base [H + ] [A - ] K a = ––––––––– = acid dissociation constant [HA] R-C-OH R-C-O - + H + (C-term/Asp/Glu) R-NH 3 + R-NH 2 + H + (N-term/Lys) HA A - + H + (general) O O

Weak acids and weak bases + H + HA A-A- K eq = K a = [H + ] [A - ] [HA] = 1.7 x 10 -5 M pK a = -log(K a ) = -log(1.7 x 10 -5 M) = 4.8 Acetic acid is a weak acid as it does not completely dissociate in water. - [H + ] [OH - ] [H 2 O] = 1.8 x 10 -16 M with [H 2 O] = 55.6 M! K eq = Recall for water:

K a = [A - ][H + ] [HA] when [A - ] and [HA] are equal then K a = [H + ]. And thus pK a = 4.8 = -log(K a ) = -log [H + ] = pH (since pH is defined as -log [H + ]) pK a and pH

pH = pK a + log [A - ] [HA] K a = [H + ] [A - ] [HA] Start at low pH and begin to add HO -. The product of [H + ] [HO - ] must remain constant, so adding HO - means [H + ] must decrease and thus pH increases. At the pK a, [A - ] and [HA] are equal, so adding more HO - does not change the ratio of [A - ] to [HA] very much and thus the pH does not change very much (shallow slope of titration curve from ~1 pH unit below pK a to ~1 pH unit above). Titration curves +/- 1 pH unit

Measuring pK a values NH 4 + H + + NH 3 [NH 4 + ] [H + ] [NH 3 ] K a = pK a = pH when [NH 4 + ] = [NH 3 ] -

K a = [H + ] [A - ] [HA] take the -log on both sides The Henderson-Hasselbalch Equation -log K a = -log [H + ] -log [A - ] [HA] pH =pK a + log [A - ] [HA] = pK a + log [Proton acceptor] [Proton donor] HA H + + A - pK a = pH -log [A - ] [HA] Apply definition p(x) = -log(x) and finally solve for pH…

Acetic acid has a pK a of 4.8. How many ml of 0.1 M acetic acid and 0.1 M sodium acetate are required to prepare 1 liter of 0.1 M buffer with a pH of 5.8? Substitute the values for the pK a and pH into the Henderson-Hasselbalch equation: 5.8 = 4.8 + log [Acetate] [Acetic acid] 1.0 = log [Acetate] 10 x then *[Acetic acid] [Acetic acid] 10 [Acetic acid] = [Acetate] For each volume of acetic acid, 10 volumes of acetate must be added (total of 11 volumes). Acetic acid needed: 1/11 x 1,000 ml = 91 ml Acetate needed: 10/11 x 1,000 ml = 909 ml on both sides

At the pK a, [HA] = [A - ] so the system is able to absorb the addition of HO - or H +. If we add HO - near the pH where [HA] = [A - ] (ie pH ~= pK a ) then HA can release H + to offset the HO - added but the ratio of HA to A - does not change much. If we add H + then A - can absorb H + to form HA. Hence, the pH does not change much. How does a buffer work? -

Buffers are vitally important in biochemical systems since pH needs to be controlled. Living systems must be “buffered” to resist large variations in pH. Phosphate buffering H 3 PO 4 H + + H 2 PO 4 - pK a1 = 2.2 H 2 PO 4 - H + + HPO 4 2- pK a2 = 7.2 HPO 4 2- H + + PO 4 3- pK a3 = 12.7 Carbonate buffering CO 2 + H 2 O H 2 CO 3 H 2 CO 3 H + + HCO 3 - pK a1 = 6.4 HCO 3 - H + + CO 3 2- pK a2 = 10.2

Phosphate buffering

CO 2 + H 2 O H 2 CO 3 H 2 CO 3 H + + HCO 3 - pK a = 6.4 HCO 3 - CO 3 -2 + H + pK a = 10.2 (not relevant, far from pH 7.4) Carbon dioxide - carbonic acid - bicarbonate buffer

What happens to blood pH when you hyperventilate?hyperventilate What happens to blood pH when you hypoventilate? If blood pH drops due to metabolic production of H + then [H 2 CO 3 ] increases by protonation of HCO 3 -, H 2 CO 3 rapidly loses water to form CO 2 (aq), which is expelled as CO 2 (g). If the blood pH rises, [HCO 3 - ] increases by deprotonation of H 2 CO 3, then breathing rate changes and CO 2 (g) is converted to CO 2 (aq) and then to H 2 CO 3 in the capillaries in the lungs.

What is the pH of 0.15 M acetic acid? The pK a of acetate is 4.8, so the K a = 10 -4.8 M = 1.58 x 10 -5 M. [H + ] [A - ] K a = _________ [HA] [H + ] 2 [H + ] 2 K a = ––––– = ––––––––– = 1.58 x 10 -5 M [HA] 0.15 M - [H + ] [H + ] 2 +1.58 x 10 -5 M [H + ] + (-2.37 x 10 -6 M 2 ) = 0 (ax 2 +bx+c = 0)(ax 2 +bx+c = 0) [H + ] = 1.53 x 10 -3 M and thus pH = 2.8 R-C-OH R-C-O - + H + O O and [H + ]=[A - ] [HA]=0.15-[H + ] ax 2 + b x + c = 0 Quadratic Formula Quadratic Formula

What is the pH of 0.15 M acetic acid? The pK a of acetate is 4.8, so the K a = 10 -4.8 M = 1.58 x 10 -5 M. [H + ] [A - ] K a = _________ [HA] [H + ] 2 [H + ] 2 K a = ––––– = ––––––––– = 1.58 x 10 -5 M [HA] 0.15 M - [H + ] Assumption: [H+] << 0.15 M! [H + ] 2 = 0.15 M * 1.58 x 10 -5 M [H + ] = 1.54 x 10 -3 M or 0.00154 M and thus pH = 2.8 R-C-OH R-C-O - + H + O O and [H + ]=[A - ] [HA]=0.15-[H + ] [H + ] 2 = 2.37x10 -6 M 2 Assumption: [H+] << 0.15 M!

Your 199 prof asks you to make a pH 7 phosphate buffer. You already have 0.1 M KH 2 PO 4. What concentration of K 2 HPO 4 do you need? pH = 7 = pK a + log 7 = 7.2 + log(x / 0.1 M) -0.2 = log(x / 0.1 M) 10 -0.2 = x / 0.1 M x = 0.063 M = [K 2 HPO 4 ] [HPO 4 2- ] [H 2 PO 4 - ] H 2 PO 4 - HPO 4 2- + H + pK a = 7.2 KH 2 PO 4 H 2 PO 4 - + K + and K 2 HPO 4 HPO 4 2- + 2 K +

Phosphate buffering

Make 200 ml of 0.1 M Na acetate buffer pH 5.1, starting with 5.0 M acetic acid and 1.0 M NaOH. Strategy 1.Calculate the total amount of acetic acid needed. 2. Calculate the ratio of the two forms of acetate (A - and HA) that will exist when the pH is 5.1. 3.Use this ratio to calculate the % of acetate that will be in the A - form. 4.Assume that each NaOH will convert one HAc to an Ac -. Use this plus the % A - to calculate the amount of NaOH needed to convert the correct amount of HAc to Ac -. Another HH calculation

(1) How much acetic acid is needed? 200 ml x 0.1 mol/l = 200 ml x 0.1 mmol/ml = 20 mmol 5.0 mol/l x x ml = 5.0 mmol/ml x x ml = 20 mmol x = 4.0 ml of 5.0 M acetic acid are 20 mmol (2) What is the ratio of Ac - to HAc at pH 5.1? 5.10 - 4.76 = log[A - ]/[HAc], thus [Ac - ]/[HAc] = 2.19 / 1 (3) What fraction of total acetate is Ac - at pH 5.1? [Ac - ] [Ac - ] 2.19 –––– = 2.19; –––––––––– = ––––––– = 0.687 or 68.7% [HAc] [HAc] + [Ac - ] 1 + 2.19 pH =pK a + log HH equation [Ac - ] [HAc]

(4) How much OH - is needed to obtain 68.7% Ac - ? Na + + OH - + HAc  Na + + Ac - + H 2 O mmol NaOH = 0.687 x 20 mmol = 13.7 mmol 1.0 mol/l x x ml = 1.0 mmol/ml x x ml = 13.7 mmol x = 13.7 ml of 1.0 M NaOH (5) Final answer (Jeopardy…) 4.0 ml of 5.0 M acetic acid 13.7 ml of 1.0 M NaOH Bring to final volume of 200 ml with water (ie add about 182.3 ml of H 2 O).

Download ppt "Bio 98 - Lecture 2 Acid-Base Equilibria, pH and Buffers."

Similar presentations