PURPOSE: 1. To understand the properties of buffer solutions. 2. To calculate the ph of buffer solutions and compare the calculated values with the experimentally determined ph values. PRINCIPLES: I. Definition, Properties and Effectiveness of Buffer Solutions A buffer solution is an aqueous solution that resists changes in ph upon the addition of small amounts of acids and bases. In order for the solution to resist changes in ph, a weak acid (HA) and its conjugate base (A ), which are the buffering species, must be within a factor of 10 of each other in concentration. This range assures that there is an appreciable amount of conjugate base (A ) to react with any added acid and that there is an appreciable amount of the weak acid (HA) to react with any added base. The following summarizes information that is essential to understanding the properties of buffer solutions: 1. A weak acid by itself [HC 2 H 3 O 2 (aq)], even though it ionizes to form some of its conjugate base [C 2 H 3 O 2 - (aq)], does not contain sufficient base [C 2 H 3 O 2 - (aq)] to be a buffer. If acid is added, there is too little conjugate base [C 2 H 3 O 2 - (aq)], to keep the ph constant. Similarly, a weak base by itself [NH 3 (aq)], even though it partially ionizes in water to form some of its conjugate acid [NH 4 (aq)], does not contain sufficient acid to be a buffer If base is added, there is too little conjugate acid [NH 4 (aq)], to keep the ph constant. 2. Solutions containing comparable amounts of both the weak acid, such as [HC 2 H 3 O 2 (aq)], - and its conjugate base, such as [C 2 H 3 O 2 (aq)], can act as buffers. These solutions have buffer capacity. The [HC 2 H 3 O 2 (aq)] present can react with small amounts of added base - and the [C 2 H 3 O 2 (aq)] present can react with small amounts of added acid. - The weak acid [HC 2 H 3 O 2 (aq)], and its conjugate base [C 2 H 3 O 2 (aq)], work together to keep the ph relatively constant in the following manner: The weak acid [HC 2 H 3 O 2 (aq)] neutralizes added base: HC 2 H 3 O 2 (aq) NaOH(aq) NaC 2 H 3 O 2 (aq) H 2 O(l) The conjugate base [C 2 H 3 O 2 - (aq)] neutralizes added acid: C 2 H 3 O 2 - (aq)] HCl(aq) HC 2 H 3 O 2 (aq) H 2 O(l) 3. The concentration of the conjugate base, [A ] and of the weak acid [HA] must be reasonably close to each other (must be within a factor of 10 of each other). [Base] The closer the Ratio of to 1, the more effective the buffer. [Acid] This implies that if in a buffer solution: [Base] [Acid], the buffer is able to perform its double duty very well in maintaining its ph almost constant when either an acid or a base is added to the buffered solution. Fall 2015/www.proffenyes.com 1
To sum up: An effective buffer must have a [Base]/[Conjugate Acid] Ratio in the range of 0.10 to 10. A buffer is most effective (most resistant to ph changes upon the addition of small to moderate amounts of either an acid or a base) when: the concentration of base and its conjugate acid are equal. the [Base]/[Conjugate Acid] Ratio is closest to 1 II. Calculating the ph of Buffer Solutions Consider a buffer solution that initially contains HC 2 H 3 O 2 (aq) and NaC 2 H 3 O 2 (aq), each at a concentration of 0.100 M. The acetic acid ionizes according to the following reaction: HC 2 H 3 O 2 (aq) H 2 O(l) H 3 O (aq) C 2 H 3 O - 2 (aq) Initial Concentrations: 0.100 M 0.100 M Note that this equilibrium reaction is different from the familiar acid ionization equilibrium of acetic acid that initially does not contain any C 2 H 3 O 2 - ions. The initial presence of the C 2 H 3 O 2 - in the equilibrium reaction given above acts as a stress on the equilibrium system and causes the equilibrium system to shift to the left. This causes the HC 2 H 3 O 2 (aq) to ionize much less that it normally would, resulting in a less acidic solution whose [H 3 O ] ion concentration is practically negligible. Keeping this in mind, the ph of this buffer solution can be calculated by using two different methods: 1. The Equilibrium Method (KHC 2 H 3 O 2 = 1.8 x 10-5 ) HC 2 H 3 O 2 (aq) H 2 O(l) H 3 O (aq) C 2 H 3 O - 2 (aq) Initial: 0.100 0.00 0.100 M Change: - x x x Equilibrium: 0.100 x x 0.100 x [H 3 O ][C 2 H 3 O - 2 ] x (0.100 x) Ka = 1.8 x 10-5 = = [HC 2 H 3 O 2 ] 0.100 - x Assuming that x is small yields: x (0.100) 1.8 x 10-5 = x = [H 3 O ] = 1.8 x 10-5 0.100 We need to confirm that the assumption that x is indeed small enough to allow the approximation used in this calculation. This confirmation is done by calculating the ratio of x and the number it was subtracted from (or added to) in the approximation. This ratio should be less than 0.05 (or 5%) Fall 2015/www.proffenyes.com 2
1.8 x 10-5 0.100 X 100 = 0.018 % < 5% The approximation is valid! x = [H 3 O ] = 1.8 x 10-5 ph = - log [1.8 x 10-5 ] = 4.74 2. The Henderson-Hasselbach equation (KHC 2 H 3 O 2 = 1.8 x 10-5 ) The calculation of the ph of a buffer solution can be simplified by relating the ph of the buffer solution to the initial concentrations of the buffer components. Consider a buffer containing the weak acid HA and its conjugate base, A - The weak acid ionizes as follows: HA(aq) H 2 O(l) H 3 O (aq) A - (aq) [H 3 O ][A - ] [HA] Ka = Solving for [H 3 O ] yields: [H 3 O ] = Ka [HA] [A - ] We can derive an equation for the ph of any buffer solution by taking the negative logarithm of both sides of the equation. [HA] - log [H 3 O ] = - log Ka [A - ] Rearrangement of this equation yields: [A - ] - log [H 3 O ] = - log Ka log [HA] Note that: - log [H 3 O ] = ph and - log Ka = pka The equation becomes: ph = pka log [A - ] [HA] The more general form of this equation, known as the Henderson-Hasselbach equation is: ph = pka log [base] [acid] The Henderson Hasselbach equation can be used to calculate the ph of any buffer solution from the initial concentrations of the buffer components as long as the x is small approximation is valid. Fall 2015/www.proffenyes.com 3
PROCEDURE PART I: Preparation of Solutions 1. The molarity of all the reagent solutions provided is 0.10 M 2. Measure the volumes of sodium hydroxide solution, acetic acid solution and deionized water into six separate labeled vials, according to the table below. The solutions can be dispensed from the burets set up for this purpose in the lab. Note that he solutions in Table A and Table B are duplicates of the same solutions, but are labeled differently. TABLE A Solution # HC 2 H 3 O 2 (aq) 0.10 M NaOH(aq) 0.10 M H 2 O(l) 1A 10.00 0.00 10.00 2A 10.00 5.00 5.00 3A 10.00 7.50 2.50 TABLE B Solution # HC 2 H 3 O 2 NaOH H 2 O 1B 10.00 0.00 10.00 2B 10.00 5.00 5.00 3B 10.00 7.50 2.50 3. Tightly seal each vial with a # 4 rubber stopper. 4. Mix the contents of each vial by slowly inverting the vials several times. Do not spill any of the contents of the vial. If you do, discard the solution and make a new solution. 5. Place the six vials in the test tube rack. PART II: Measurement of ph 1. Calibrate your ph meter for buffer ph = 7.01, followed by calibration for buffer ph = 4.01 2. Measure the ph of each of your six solutions (1A, 2A, 3A and 1B, 2B and 3B) and record these measurements. 3. Turn the ph meter OFF 4. Rinse the electrode with a stream of deionized water and catch the water in a beaker. 5. Remove the excess water from the electrode with tissue paper. 6. Store the electrode temporarily in the ph = 7.01 buffer solution. 7. Calculate the average ph for each solution (1, 2 & 3) Fall 2015/www.proffenyes.com 4
PART III: Determination of Buffering Effectiveness 1. Addition of 0.10 M hydrochloric acid to solutions 1A, 2A and 3A Add 2.00 ml of 0.10 M hydrochloric Acid (HCl) to solutions 1A, 2A and 3A by dispensing it from the buret, set up for this purpose in the lab. Cap or seal the vials containing the three solutions. Mix the contents of each tube vial by inverting the tubes vials several times. Measure the ph of each of your three solutions, to which hydrochloric acid has been added (1A Acid), (2A Acid) and (3A Acid). Record these measurements. Turn the ph meter OFF Rinse the electrode with a stream of deionized water and catch the water in a beaker. Remove the excess water from the electrode with tissue paper. Store the electrode temporarily in the ph = 7.01 buffer solution. 2. Addition of 0.10 M sodium Hydroxide to solutions 1B, 2B and 3B Add 2.00 ml of 0.10 M sodium hydroxide to solutions 1B, 2B and 3B by dispensing it from the buret, set up for this purpose in the lab. Cap or seal the vials containing the three solutions. Mix the contents of each vial) by inverting the vials several times. Measure the ph of each of your three solutions, to which sodium hydroxide has been added (1B Base), (2B Base) and (3B Base). Record these measurements. Turn the ph meter OFF Rinse the electrode with a stream of deionized water and catch the water in a beaker. Remove the excess water from the electrode with tissue paper. 3. Storing the ph-meter after use. Make sure the ph meter is turned OFF! Rinse the electrode with a stream of deionized water over the sink or catch the water in a beaker. Remove the excess water from the electrode with tissue paper (Kimwipe) When finished, place the ph meter in the common plastic bucket containing the ph = 7.01 buffer solution. 4. Storing the two buffer solutions : DO NOT DISCARD THE BUFFER SOLUTIONS! The plastic test-tubes containing the two buffer solutions should be kept in your locker (capped) for the next experiment. CALCULATIONS PART I & PART II [Base] Calculate the Ratio for solutions 1, 2 & 3 [Acid] 1. Recall that solutions 1A and 1B, 2A and 2B, 3A and 3B respectively, are initially identical 2. Calculate the average initial ph obtained from two measurements for each solution Fall 2015/www.proffenyes.com 5
3. To calculate the [BASE]/[ACID] Ratio: Solution 1: The calculation of the [BASE]/[ACID] Ratio is an equilibrium calculation in which we use the concentration of the [H ] obtained from a ph measurement of solution 1. An Equilibrium Table (ICE Table) will be used for this calculation. Solutions 2 & 3: The calculation of the [BASE]/[ACID] Ratio is based on a stoichiometry calculation involving a limiting reagent. PART III: Buffering Effectiveness 1. Determine the change in ph for solutions 1A, 2A, and 3A, when 2.00 ml of 0.10 M of hydrochloric acid is added ( ph acid ) 2. Determine the change in ph for solutions 1B, 2B, and 3B, when 2.00 ml of 0.10 M of sodium hydroxide is added ( ph base ) 3. Average ( ph acid ) and ( ph base ) for solutions: 1A &1B and 2A & 2B and 3A & 3B, respectively. 4. Evaluate the relative buffer effectiveness of the three solutions: Based on experimental evidence (the Average Change in ph) Based on the respective [BASE]/[ACID] Ratio. Provide the reason that accounts for the fact that: One solution cannot reasonably act as a buffer, and One solution is a more effective buffer than another solution. Part IV: Error Analysis 1. Calculate the initial ph of the two buffer solutions: By using the Equilibrium method, and By using the Henderson Hasselbach equation 2. Compare the calculated ph values of the two buffer solutions with the experimentally determined ph values. Bibliography: 1. Nivaldo J. Tro, Chemistry: A Molecular Approach, Third Edition Fall 2015/www.proffenyes.com 6
REPORT FORM NAME: Date: Partner: Part I & Part II: Preparation of Solutions and ph Measurement of Solutions Average ph of solutions Solution 1A 1B 2A 2B 3A 3B Measured ph Average ph 1. ADDITION OF HYDROCHLORIC ACID Molarity of Hydrochloric Acid: Volume of Hydrochloric Acid added M ml ph after addition of hydrochloric acid: Solution 1A Hydrochloric Acid: ph = Solution 2A Hydrochloric Acid: ph = Solution 3A Hydrochloric Acid: ph = 2. ADDITION OF SODIUM HYDROXIDE Molarity of Sodium Hydroxide: Volume of Sodium Hydroxide solution added: M ml ph after addition of sodium hydroxide: Solution 1B Sodium Hydroxide: ph = Solution 2B Sodium Hydroxide: ph = Solution 3B Sodium Hydroxide: ph = Fall 2015/www.proffenyes.com 7
3. Calculate the Base/Acid Ratio for Solution 1: Original Molarity of HC 2 H 3 O 2 : M Molarity of HC 2 H 3 O 2 solution before mixing (M) Volume of HC 2 H 3 O 2 solution added Volume of water added Total Volume of Solution 1 Molarity of HC 2 H 3 O 2 in Solution 1 M Average Measured ph of Solution 1 Average [H ] in Solution 1 M Complete the Equilibrium Table below for Solution 1: DO NOT USE X! Use the Actual Values indicating the changes in concentrations. HC 2 H 3 O 2 (aq) H (aq) C 2 H 3 O 2 (aq) Initial: Change Equilibrium: Calculate the Ratio: [C 2 H 3 O 2 ] [H C 2 H 3 O 2 ] to the proper amount of significant figures. Fall 2015/www.proffenyes.com 8
4. Calculate the Base/Acid Ratio for Solution 2: Original Molarity of HC 2 H 3 O 2 : M Table II A Molarity Volume of Number of of HC 2 H 3 O 2 moles of HC 2 H 3 O 2 solution HC 2 H 3 O 2 solution added added before mixing (M) Molarity of NaOH: M Table II B Molarity Volume of Number of of NaOH moles of NaOH solution NaOH solution added added before mixing (M) Balanced Chemical Equation that illustrates the reaction that takes place in test tube/vial: H 2 O(l) Net Ionic Equation: H 2 O(l) Start: mol mol mol End: mol mol mol What is the total volume of solution after mixing? ml Calculate the molarity of [HC 2 H 3 O 2 ] after mixing: M (show calculations below) Calculate the molarity of [C 2 H 3 O 2 ] after mixing: M (show calculations below) Calculate the Ratio: [C 2 H 3 O 2 ] to the proper amount of significant figures. [H C 2 H 3 O 2 ] Fall 2015/www.proffenyes.com 9
5. Calculate the Base/Acid Ratio for Solution 3: Original Molarity of HC 2 H 3 O 2 : M Table III A Molarity Volume of Number of of HC 2 H 3 O 2 moles of HC 2 H 3 O 2 solution HC 2 H 3 O 2 solution added added before mixing (M) Molarity of NaOH solution before mixing (M) Molarity of NaOH: M Table III B Volume of Number of NaOH moles of solution NaOH added added Balanced Chemical Equation that illustrates the reaction that takes place in test tube: H 2 O(l) Net Ionic Equation: H 2 O(l) Start: mol mol mol End: mol mol mol What is the total volume of solution after mixing? ml Calculate the molarity of [HC 2 H 3 O 2 ] after mixing: M (show calculations below) Calculate the molarity of [C 2 H 3 O 2 ] after mixing: M (show calculations below) Calculate the Ratio: [C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] to the proper amount of significant figures. Fall 2015/www.proffenyes.com 10
Part III. Comparing the buffering effectiveness of the three solutions TABLE IV A: Change in ph upon addition of acid ( phacid) Initial Average ph ph after the addition of 2.00 ml of 0.10 M HCl Change in ph ( ph acid) Solution 1 Solution 2 Solution 3 5.00 ml NaOH 5.00 ml H 2 O 10.00 ml H 2 O 7.50 ml NaOH 2.50 ml H 2 O TABLE IV B: Change in ph upon addition of base ( phbase) Solution 1 Solution 2 Solution 3 10.00 ml H 2 O 5.00 ml NaOH 5.00 ml H 2 O 7.50 ml NaOH 2.50 ml H 2 O Initial Average ph ph after the addition of 2.00 ml of 0.10 M NaOH Change in ph ( ph base) TABLE IV C: Summary of change in ph upon addition of acid or base (transfer and average your data from Table IV A and IV B above) Solution 1 Solution 2 Solution 3 Change in ph ( phacid) Change in ph ( phbase) Average change in ph: ( phacid phbase) 2 [BASE] Ratio [ACID] 10.00 ml H 2 O 5.00 ml NaOH 5.00 ml H 2 O 7.50 ml NaOH 2.50 ml H 2 O Fall 2015/www.proffenyes.com 11
Questions: 1. Which solution cannot be considered an effective buffer? (a) (b) What is the experimental evidence that supports your answer in (a) above? (c) What is the reason that this solution cannot be considered an effective buffer? 2. Which solutions can be considered is effective buffers? (a) and (b) What is the experimental evidence that supports your answer in (a) above? (c) What is the reason that these solutions is are effective buffers? 3. (a) Which of the solutions listed in # 8. above is the most Effective Buffer? (b) What is the experimental evidence that supports your answer in (a) above? (c) What is the reason that this solution is the most Effective Buffer? Fall 2015/www.proffenyes.com 12
Part IV: Error Analysis Calculate the theoretical initial ph of the two buffer solutions: Solution 2: Concentration of components: Molarity of [HC 2 H 3 O 2 ] M Molarity of [C 2 H 3 O 2 ] M a. Use the Equilibrium Method (you may use the approximation method) Equation: Initial: Change: Equilibrium: Show Calculations below: [H 3 0 ] = Check if the approximation is valid. Show calculations below: ph = b. Use the Henderson Hasselbach equation Show calculations below: ph = Fall 2015/www.proffenyes.com 13
Solution 3: Concentration of components: Molarity of [HC 2 H 3 O 2 ] M Molarity of [C 2 H 3 O 2 ] M a. Use the Equilibrium Method (you may use the approximation method) Equation: Initial: Change: Equilibrium: Show Calculations below: [H 3 0 ] = Check if the approximation is valid. Show calculations below: ph = b. Use the Henderson Hasselbach equation Show calculations below: ph = Fall 2015/www.proffenyes.com 14
SUMMARY Buffer Solutions Solution 2 Calculated Values ph calculated by Equilibrium Method ph calculated by using the Henderson- Hasselbach equation Experimentally measured ph values % Error Solution 3 Fall 2015/www.proffenyes.com 15
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