Buffers, Electrochemistry Practical Lesson on Medical Chemistry and Biochemistry General Medicine Jan Pláteník & Tomáš Navrátil 2010/2011 1 BUFFERS AND BUFFER CAPACITY 1.1 Principle of buffering: A buffer solution is a solution that resists changes in ph either when diluted or when limited amounts of acid or base are added to it. Such a solution can be prepared by combining a weak acid and its salt with a strong base (conjugated base) or, analogously, a weak base and its salt with a strong acid (conjugated acid). For example: Acetate buffer: CH 3 COOH (the weak acid) + CH 3 COONa (the salt, conjugated base) Phosphate buffer: NaH 2 PO 4 (the weak acid) + Na 2 HPO 4 (the salt, conjugated base) Tris buffer: (Tris: Tris [2-amino-2-(hydroxymethyl)-propan-1,3-diol)], an organic base) 1
The Henderson-Hasselbalch equation describes the behaviour of such a buffer and for the mixture of a weak acid and its salt with a strong base (conjugated base) it has the form: ph = pk a + cs log cac pk a negative logarithm of the dissociation constant for the weak acid, c s substance concentration of the salt (conjugated base), substance concentration of the weak acid (conjugated acid). c ac Graphically the Henderson-Hasselbalch equation plotted as the acid : conjugated base ratio vs. ph of buffer actually constitutes the titration curve of the weak acid (see figure on this page). Note also that for the acid : base ratio 1:1 the ph of buffer just equals the pk a (this is valid for uni-univalent systems, e.g., CH 3 COOH/CH 3 COONa, NaH 2 PO 4 /Na 2 HPO 4 ). 10 Titration curve of sodium phosphate buffer 9 ph of buffer 8 7 6 ph = pka = 7.21 5 9:1 8:2 7:3 6:4 1:1 4:6 3:7 2:8 1:9 Acid : base ratio The equation for a weak base and its salt with a strong acid (conjugated acid) has the form: ph = pkw pkb + cb log cs pk b negative logarithm of the dissociation constant for the weak base, c b substance concentration of the base, c s substance concentration of the salt (conjugated acid), pk w = 14 = log 10-14 (ionic product of water). 2
Buffer capacity (β) is defined as the amount of a strong acid or a strong base that has to be added to 1 litre of a buffer to cause ph change of 1.0 ph unit: β = cb cac = ph ph The buffer capacity depends on the amounts of substance of the weak acid and its conjugated base in the buffer. It is in fact directly related to the first derivative of the buffer titration curve, or, in other words, the slope of the titration curve. As the slope of the titration curve is the smallest at the acid : base ratio 1:1, the buffer capacity is maximal at the same point (the second derivative of the titration curve is equal to zero at the same point). 1.2 Calculation of ph of buffer after addition of strong acid or base Example: Let s have 600 ml of a sodium phosphate buffer of concentration c = 0.25 mol/l and the ratio acid : base is 2:3. Next, we add 150 ml of HCl, c = 0.2 mol/l. How much the ph changes after this addition? (pk a of sodium dihydrogen phosphate is 7.21) Solution: In order to calculate the ph of this buffer before and after addition of HCl, we need to know the ratio of acid : base for both conditions. The ratio is given for the original buffer as 2:3, so we can easily use the Henderson-Hasselbalch equation to calculate the original ph: ph = pk a + log 3/2 = 7.21 + log 1.5 = 7.21 + 0.17609 = 7.39. Next, we need to consider how the ratio acid : base changes following addition of HCl. The strong acid actually titrates the basic component of the buffer, converting it into the acidic component: HCl + Na 2 HPO 4 NaH 2 PO 4 + NaCl before addition of HCl: Number of moles for total phosphate: (0.6 x 0.25) = 0.15 mol NaH 2 PO 4 : 0.15 /5 x 2 = 0.06 mol Na 2 HPO 4 : 0.15 /5 x 3 = 0.09 mol after addition of HCl 0.15 x 0.2 = 0.03 mol: Number of moles for total phosphate: (0.6 x 0.25) = 0.15 mol NaH 2 PO 4 : 0.06 mol + 0.03 mol = 0.09 mol Na 2 HPO 4 : 0.09 mol 0.03 mol = 0.06 mol The new ratio of acid : base is 0.09 : 0.06, i.e. 3 : 2. And the resulting ph is: ph = pk a + log 2/3 = 7.21 + log 0.6667 = 7.21 0.17609 = 7.03. The ph change (effect of addition): ph = 7.39 7.03 = 0.36 3
1.3 Experiment I: Buffer capacity Reagents: 1. NaH 2 PO 4 0.1 mol/l 2. Na 2 HPO 4 0.1 mol/l 3. NaCl 0.1 mol/l 4. HCl 0.1 mol/l 5. NaOH standard solution 0.1 mol/l Procedure: a. Use the pre-marked plastic containers for ph measurements and prepare the initial mixtures 1-4 as shown in the following table: Container No 1 2 3 4 Na 2 HPO 4 0.1 mol/l ml 5 1 9 - NaH 2 PO 4 0.1 mol/l ml 5 9 1 - NaCl 0.1 mol/l ml - - - 10 acid/conjugate base ratio ph measured ph calculated Add to all tubes ml ml ml acid or base 0.1 mol/l 1.0 1.0 1.0 1.0 ph measured ph calculated 1 b. Using the Henderson-Hasselbalch equation, calculate the ph values of prepared phosphate buffers (pk a = 7.21) before as well as after addition of 1 ml of HCl 0.1 mol/l. c. Measure ph of the initial mixtures and compare these values with the calculated ones. d. Then add to each of the four tubes 1 ml of HCl 0.1 mol/l, mix and measure their ph values again. Compare the obtained values with calculated ones. e. Conclude which of the prepared solutions shows the highest buffer capacity. Do all the examined solutions behave as buffers? f. Alternatively, you can test the buffer capacity of the prepared solutions by adding 1 ml of NaOH 0.1 mol/l instead of HCl. 1 Only in cases when the buffering capacity has not been exceeded (over-titrated) 4
1.4 Experiment II: Effect of ionic strength on ph of buffer Reagents: 1. NaH 2 PO 4 0.1 mol/l 2. Na 2 HPO 4 0.1 mol/l Procedure: a. Prepare a stock phosphate buffer for this experiment: to a glass beaker measure 10 ml of NaH 2 PO 4 0.1 mol/l, add 10 ml of Na 2 HPO 4 0.1 mol/l, and mix. b. To the pre-marked plastic containers (rinse them well after the previous experiment) prepare the working buffers 1-4 according to the table: Container No. 1 2 3 4 Stock phosphate buffer 10 ml 4 ml 1 ml 1 ml from container No. 3 Distilled water 0 6 ml 9 ml 9 ml Concentration of buffer (mol/l) 0.1 0.04 0.01 0.001 Acid:base ratio 1:1 1:1 1:1 1:1 ph measured ph calculated/predicted c. Measure ph of all the mixtures and record the values to the table. d. Use the Henderson-Hasselbalch equation to calculate the ph values of prepared phosphate buffers (pk a = 7.21). e. Compare the measured values with the predicted ones and try to explain the difference. 5
2 Electrochemical sources of electric potential 2.1 Principle Let s consider a metallic, e.g. Zn rod that is placed into solution of an electrolyte, such as NaCl. Although zinc is not significantly soluble in aqueous NaCl, atoms of zinc still tend to enter the solution as its cations Zn 2+, leaving their electrons in the rod. At an equilibrium state an electric bilayer is developed and the metal rod is negatively charged with respect to the surrounding solution. The rod (electrode) gains an electric potential called an electrode potential. It is also called a redox potential, since dissolution of the metal is accompanied by its oxidation (loss of electrons) while the opposite process would be reduction (gain of electrons). This structure is called a half cell. Now let s consider that to the vessel with NaCl and Zn electrode we add another rod from a different metal, e.g., copper. It would also dissolve and form cations Cu 2+, but not so much as zinc rod because in comparison to zinc, copper exhibits a higher redox potential (i.e., it more attracts electrons). The zinc rod (electrode) becomes negatively charged with respect to the copper one. Finally, let s connect the two rods with a metallic wire: electrons from the more negative zinc rod can now flow to the positive copper electrode. In other words, an electric current is generated we have created a galvanic cell. Much more electricity is generated if such a galvanic (voltaic cell) consists from two halfcells, first one containing Zn in solution of Zn 2+, while the second made from Cu in solution of Cu 2+. If the cells are connected with a salt bridge that allows migration of ions but not bulk mixing of the two solutions, and the rods are connected with a metallic wire, the overall oxidation-reduction (redox) reaction between Zn and Cu Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) is separated in two half-reactions. At the Zn electrode (anode) oxidation occurs Zn(s) Zn 2+ (aq) + 2 e while at the copper rod (cathode) reduction takes place Cu 2+ (aq) + 2 e Cu(s) Electrons must in this setting flow through the external wire connection between the electrodes; the difference in redox potential between the two redox couples is the driving force that is used to generate electricity. Accumulators and batteries widely used nowadays utilize various compounds in various settings, but all of them operate on this principle. In this experiment we prove and examine formation of a galvanic cell between various metals, immersed in the physiological saline. The resulting potential is measured with a sensitive voltmeter. Before measurement, consider values of standard redox potentials for the metals involved: 6
Standard redox potentials Half-cell E o (V) Na/Na + 3.02 Al/Al 3+ 1.66 Zn/Zn 2+ 0.76 Fe/Fe 2+ 0.44 Sn/Sn 2+ 0.14 Pb/Pb 2+ 0.13 H 2 /H + 0.00 Cu/Cu 2+ + 0.34 2+ 2Hg/Hg 2 + 0.80 Ag/Ag + + 0.81 Pt/Pt 2+ + 1.20 Au/Au 3+ + 1.42 2.2 Experimental Reagents and equipment/tools: 1. Analog voltmeter HD-075 1V 2. Petri dish with physiological saline 3. Electrodes from different metals (Pt, Fe, Cu, Ag, Sn, and reference Al) 4. Wires with connectors 5. Crocodile clips 6. Beaker with concentrated HCl Procedure: a. Attach the crocodile clips to both sides of the wires with connectors. b. Clip both wires to the appropriate contacts on the rear of the voltmeter: one to the - contact and the other one to the +1 sign (see also figure and notes below). c. Attach the reference electrode (piece of aluminium foil) to the other end of the - wire. d. Attach a positive electrode (one of the tested metals) to the clip on the +1 wire. e. Clean the piece of metal serving as a positive electrode: dip it briefly to the beaker with concentrated HCl (dissolves oxide coating on the tested metal), wash with distilled water from squeeze bottle and wipe with piece of tissue. f. The actual measurement is best done in pairs. One person immerses both the examined electrodes into the saline solution in the dish so that only the tested metals, not the clips themselves (!), are immersed. The electrodes must not directly touch each other. The other person simultaneously reads the electric potential on the voltmeter scale. The highest transient value is the one to be recorded. g. Repeat each measurement three times and calculate arithmetic mean of the three values obtained. h. Perform the same for all other combinations of metals available as directed by the table below. 7
i. Compare your experimental data with the known differences in standard redox potentials summarized above. Electrolyte: saline Electrochemical potential E [V] Electrode 1 Electrode 2 E 1 [V] E 2 [V] E 3 [V] Mean E [V] Aluminium Platinum Aluminium Iron Aluminium Copper Aluminium Silver Aluminium Tin Aluminium Aluminium Notes on measurement of electrochemical potential: The front view on the voltmeter HD-075 1 V. The scale range is 1 Volt; each small division on the scale represents 0.04 V. If the potential is read with precision to a half of the small division of the scale, the attainable precision of measurement is 0.02 V. The rear side of the voltmeter offers 5 contacts. The one marked as - is for the electrode with the more negative electrochemical potential, while the contacts marked +1, +2, +3, and +4 are for attachment of the more positive electrode. If the equipment needle goes below zero, it obviously indicates that attachment of the electrodes with respect to their polarity needs to be reversed. The measured value should be recorded with a minus sign. As a default, attach the positive electrode to the +1 connector, and then the equipment scale is 1 V. If the needle goes above 1 V, the range needs to be extended by reattachment of the positive electrode from +1 to the +2 (or even +3 or +4 ). Then the measured value should be multiplied with the number at the connector. For example, if 0.82 V is read on the scale with the positive electrode attached to +3, the result is 3 0.82 V = 2.46 V. 8