Buffers

Chemistry UNIT 4 ACIDS AND BASES PACKAGE #7 BUFFERS Buffers are solutions which resist LARGE pH changes when SMALL additions of acid or base are made to the solution. Buffers typically consist of equilibrium systems of weak acids and their conjugate base or weak bases and their conjugate acids. You need substantial amounts of the acid and its conjugate (or base and its conjugate) in order to qualify as a buffer, IN THE LAB: In order for a pH meter to be accurate, it must be standardized for the range of pH's that it is going to read. Buffer solutions can be prepared to provide a wide range of reliable pH standards. (These kinds of buffer solutions were used as standards for the pH lab that we did where we diluted HCl and CH3COOH to produce a certain pH value). By mixing equal molar quantities of a weak acid such as CH3COOH and its salt NaCH3COO ( i.e. the conjugate base of CH3COOH), you can standardize a pH meter within a certain pH range. IN NATURE / BIOLOGICAL APPLCATIONS: BLOOD: Buffers exist in human blood to maintain the pH at or near 7.4. (Slightly alkaline). Hemoglobin is the oxygen carrier in the blood and is involved in the equilibrium. HHb + O2 + H2O ⇄ H3O+ + HbO2(hemoglobin) (oxyhemoglobin) If pH is too low (called acidosis of the blood when pH is less than 7.20), which means that the [H3O+] is too high, the equilibrium shifts sufficiently left resulting in the [HbO2-] being too low. The result is that O2 will not bind with HHb properly. If pH is too high (called alkalosis of the blood when pH is > 7.50) and thus the [H3O+] is too low, then the equilibrium shifts sufficiently right resulting in the [HbO2-] being preferentially formed. The result is that the O2 is prevented from being released from the blood. To prevent alkalosis or acidosis, two buffer systems are present in our bodies to regulate the [H3O+] . Otherwise eating tomatoes or drinking lemon juice would affect the blood pH and it would be fatal! 1) The CO2 / HCO3- System A product of metabolism and respiration is CO2(g). When CO2(g) diffuses into the blood, it reacts with H2O to form carbonic acid: CO2 + H2O ⇄ H2CO3 The carbonic acid could ionize: H2CO3 + H2O ⇄ H3O+ + HCO3But if this happened, the [H3O+] of the blood would increase and the pH would go down. BUT we have an excess of HCO3- in our blood, so therefore the H2CO3 does not ionize. And breathing out CO2(g) (i.e. CO2(g) ⇄ CO2(aq) ) upsets the [H3O+] in the following reaction: CO2(aq) + H2O ⇄ H3O+ + HCO3Breathe out the CO2 and the reaction shifts left (also excess HCO3- causes shift to left) 2) The H2PO4- / HPO4-2 System Both of the above species are present in the blood and our cells (for bone, tooth, and DNA maintenance). The buffer stabilizes the pH of the cells to a large degree. H2PO4- + H2O ⇄ H3O+ + HPO4-2 As you can see, buffer solutions resist changes in pH. Buffers will either: • return the solution / sytsem to a more stable pH • produce more neutral initial solution (conjugate common ion reduces dissociation). EXAMPLES of making BUFFERS in CHEMISTRY: •ACIDIC BUFFER: MIX 1.0 mole of CH3COOH and 1.0 mole of NaCH3COO and dilute to 1.0 L of solution. When making up a buffer solution, you must ensure that the salts used are soluble. This solution is referred to as an acidic buffer because it buffers the pH in the acidic region. See the equilibrium that is established in the following sample calculations. •BASIC BUFFER: MIX 1.0 mole of NH3 and 1.0 mole of NH4NO3 and dilute to 1.0 L of solution. When making up a buffer solution, you must ensure that the salts used are soluble. This solution is referred to as an basic buffer because it buffers the pH in the basic region. The equilbrium that is established is as follows: NH4+ + H2O ⇄ NH3 + H3O+ Note that the NH4+ came from: NH4NO3 à NH4+ The NO3- is a spectator, but the NH4+ hydrolyzes + NO3- Even if you showed the basic hydrolysis of ammonia, you would still be able to calculate the pH using a Kb = Kw/Ka instead of the Ka that you would have used in the above equilibrium. NH3 + H2O ⇄ NH4+ + OH- LIMITATIONS OF BUFFERS: There is a limit to the amount of H3O+ or OH- which can be neutralized by a buffer. If there is 1 mol of the conjugate base present, then a maximum of 1 mole of H3O+ that can be neutralized. (and similarily 1 mol of conjugate acid can neutralize 1 mol of OH-). SAMPLE BUFFER CALCULATIONS: 1. Calculate the pH of an unbuffered 0.010M acetic acid solution. 2. Calculate the pH of a buffered 0.010M acetic acid solution. 3. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. 4. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. 5. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of OH- is being added from an basic source. 6. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of OH- is being added from an basic source. SAMPLE BUFFER CALCULATIONS - Answers 1. pH = 3.38 2. pH = 4.74 3. pH = 2.40 4. pH = 4.38 5. pH = 4.57 6. pH = 5.11 EXTRA BUFFER QUESTIONS: 1. Calculate the pH of a solution prepared by mixing 40.0 ml of 0.60 M HCOOH with 60.0 ml of 0.15 M HCOOK. 2. Calculate the pH of a buffer solution prepared by mixing 25.0 ml of 0.20 M acetic acid with 75.0 ml of 0.16 M sodium acetate. 3. How many grams of NaF must be added to 0.100 L of 0.20 M HF to produce a solution with a pH = 3.00 4. Find pH of a solution made by dissolving 2.0 g NH4NO3 in 50.0ml 0.400 M NH3 5. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of OHis being added from an basic source. EXTRA BUFFER QUESTIONS - Answers 1. pH = 3.32 2. pH = 5.12 3. 0.57 g 4. pH = 9.13 5. pH = 5.11 Unit 4 review list – the second ½ of Unit 4 Hydrolysis Ka Calcs Kb calcs Anhydrides Acid Rain Buffers in Chemistry calculations Buffers applications in Biology Acid-Base Neutralization calculations Titrations Titrations Curves Indicators Primary Standards SAMPLE BUFFER CALCULATIONS – FULL Answers 1. Calculate the pH of an unbuffered 0.010M acetic acid solution. CH3COOH < ====> CH3COO+ H+ I 0.010M ---------C E 0.010 -y y y Ka = 1.8 x 10-5 y = 4.2 x 10-4 M pH = 3.38 = y2 / 0.010 -y 2. Calculate the pH of a buffered 0.010M acetic acid solution. CH3COOH < ====> CH3COO+ H+ I 0.010M 0.010M ---C E 0.010 -y 0.010+ y y In a buffered sol'n, the conjugate base is also present Ka = 1.8 x 10-5 = y (0.010 + y) / 0.010 -y -5 y = 1.8 x 10 M pH = 4.74 THE PRESENCE OF THE CONJUGATE NOT ONLY PREVENTS LARGE SWINGS, IT GUARANTEES A MORE NEUTRAL INITIAL EQUILIBRIUM! 3. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. CH3COOH < ====> CH3COO+ H+ I 0.010M ------0.004 M C -y -----------> +y +y E 0.010-y y 0.004 + y Ka = 1.8 x 10-5= y (0.004 + y) / (0.010-y) y= 4.5 x 10-5 [H] = ~0.004 pH = 2.40 This is a difference of 2.34 pH units from Question #2. 4. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. CH3COOH < ====> CH3COO+ H+ I 0.010M 0.010M 0.004 M C +0.004 < ------------ -0.004 -0.004 see **** first!!! E 0.014 0.006 y~0 Ka = 1.8 x 10-5 **** K trial = (0.004)(0.0010) / (0.010) = 4 x 10-3 Ktrial > K Thus in the R line the equil. shifts left Ka = 1.8 x 10-5= y (0.006) / (0.0014) y= 4.2 x 10-5 (note that we assumed that [H] e was ~0. This assumption is valid compared to the intial concentrations). pH = 4.38 This is a difference of only 0.36 pH units from Question #2. 5. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of OH- is being added from an basic source. I C E CH3COOH 0.010M -0.004 0.006 + OH- < ====> CH3COO+ 0.004 M -------0.004 -----------> +0.004 y~0 0.004 H2O X X X K =1/Kb= Ka/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.004/ 0.006 (y) y= 3.7 x 10-10 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~3.7 x 10-10 pOH = 9.43 pH = 4.57 6. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of OH- is being added from an basic source. CH3COOH + OH- < ====> CH3COO+ H2O I 0.010M 0.004 M 0.010M X C -0.004 -0.004 -----------> +0.004 X ****** E 0.006 y~0 0.014 X ****** Ktrial = (0.010) / (0.010)(0.004) = 2.5 X 102 Ktrial < K shifts right K =1/Kb= Ka/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.014/ 0.006 (y) y= 1.3 x 10-9 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~1.3 x 10-9 pOH = 8.89 pH = 5.11 EXTRA BUFFER QUESTIONS Answers 1. 2. 3. 4. 5. I R E 3.32 5.12 0.57 g 9.13 CH3COOH 0.010M -0.004 0.006 + OH- < ====> CH3COO+ 0.004 M 0.010M -0.004 -----------> +0.004 y~0 0.014 H2O X X X ****** ****** Ktrial = (0.010) / (0.010)(0.004) = 2.5 X 102 Ktrial < Keq therefore the equilibrium shifts right K =1/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.014/ 0.006 (y) y= 1.3 x 10-9 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~1.3 x 10-9 pOH = 8.89 pH = 5.11