Buffers Lab Report

There is not a formal lab report for this lab. Complete the below pages and submit them to your TA before leaving lab. Briefly describe what a buffer is. Include the relevant conceptual background along with balanced chemical reactions to show how buffers react with a strong acid or a strong base. Introduce buffer capacity. Include a description about how the buffer region relates to the Ka of the weak acid used in making the buffer. A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. To be classified as a buffer it must be a mixture of a weak acid its conjugate base, or a weak base and its conjugate acid. Buffers are able to resist pH change through absorbing small amounts of acids (H+or H3O+ions) or bases (OH–ions) so that the pH of the system changes by a smaller amount than it would with most other solutions. Furthermore, a buffer is an equilibrium system. For example, a buffer can be prepared with nitrous acid, HNO2. The weak acid establishes an aqueous equilibrium as shown below: −¿(aq) ¿ +¿ ( aq ) +NO2 ¿ HNO2 ( aq ) ↔H The equilibrium constant (Ka) is also shown below: +¿¿ H ¿ −¿ ¿ NO2 ¿ ¿ K a=¿ Data: Insert a copy of the titration curves. One graph with an overlay of all of the titration curves is very appropriate as long as each curve is clearly labeled in the caption. Make sure to include the relevant concentrations of the components in the buffer system and the concentration of the titrant. Insert a data table with a summary of the data necessary to calculate the buffer capacity. Include a caption with the necessary detail. Graph 1: Graph displaying the four titration’s pH and Volume (mL) Table 1: Table of data collected from titration including Volume B and NaOH, volume B + NaOH 0.3 0.6 1.32 1.63 1.98 pH 4.20 4.26 4.33 4.37 4.41 volume A + NaOH 0 0.14 0.3 0.55 0.72 pH 4.24 4.44 4.58 4.80 4.92 volume B + HCl 0 0.31 0.61 0.97 1.39 pH 4.22 4.16 4.098 4.01 3.90 volume A + HCl 0 0.21 0.37 0.5 0.61 pH 4.14 3.73 3.11 2.82 2.64 2.32 2.55 2.89 3.31 3.62 3.93 4.32 4.64 4.91 5.19 5.51 5.87 6.27 6.59 6.99 7.33 7.71 10.11 10.43 10.89 11.23 11.81 12.17 12.51 12.83 13.27 13.74 14.21 14.63 15.02 15.3 15.79 16.12 16.59 17.03 17.46 17.81 17.97 18.37 18.7 19.03 19.13 19.21 19.49 19.62 19.72 4.44 4.47 4.50 4.54 4.57 4.60 4.64 4.66 4.68 4.71 4.73 4.76 4.79 4.82 4.85 4.88 4.90 5.09 5.12 5.16 5.18 5.24 5.27 5.30 5.33 5.38 5.43 5.48 5.53 5.58 5.62 5.70 5.76 5.84 5.95 6.07 6.17 6.23 6.47 6.815 10.65 11.33 11.56 11.88 11.97 12.04 0.91 1.05 1.21 1.37 1.5 1.64 1.78 1.89 2.01 2.05 2.1 2.21 2.39 2.52 2.81 3.2 3.81 4.36 5.03 5.15 5.28 5.42 5.53 5.73 5.90 6.14 6.65 7.69 11.26 11.85 12.14 12.30 12.46 12.61 12.77 12.85 1.8 2.21 2.59 2.97 3.4 3.55 3.71 4.05 4.31 4.59 4.96 3.78 3.62 3.43 3.15 2.67 2.51 2.41 2.21 2.11 2.04 1.96 0.71 0.8 0.91 1.01 1.12 1.22 1.31 1.42 1.52 1.62 1.75 1.85 2.53 2.43 2.35 2.29 2.24 2.19 2.15 2.10 2.06 2.03 2.00 1.98 20.05 20.45 20.89 21.31 21.95 22.59 12.20 12.33 12.43 12.50 12.59 12.67 Results: 1. Report the buffer capacity of Buffer A with respect to the addition of acid and the addition of base (show all calculations) Calculated valuesfor Buffer SolutionA∧NaOH: Volumeof NaOH added: 4.36mL InitialpH :4.24 FinalpH:12.85 Molesof NaOH added: 1L ( 1000mL )=0.00436LofNaOHadded 0.5mol =0.00218molof NaOH added 0.00436L× ( 1L ) 4.36mL× ( ) Buffer Capacityof A withrespect ¿ additionof NaOH= molesof NaOH added ( change ∈ pH ) BufferCapactiy= (0.00218molof NaOH added) (12.85−4.24) Buffer Capacityof A withNaOH=2.53x 10−¿ Calculated valuesfor Buffer SolutionA∧HCl: Volumeof HCl added:1.85mL InitialpH:4.14 FinalpH:1.98 Molesof HCl added: 1L =0.00185Lof NaOH added 1.85mL× 100mL 0.00185L× 0.5mol =0.000925molof NaOH added 1L ( molesof HCladded ) Buffer Capacityof A withrespect ¿ additionof HCl= ( change ∈ pH ) ( ) ( ) Buffer Capacityof A withrespect ¿ additionof NaOH=4.28×10 −4 2. Report the buffer capacity of Buffer B with respect to the addition of acid and the addition of base (show all calculations) Calculated valuesfor Buffer SolutionB∧NaOH: Volumeof NaOH added:22.29mL InitialpH:4.20 FinalpH:12.67 Molesof NaOH added: 1L =0.02229Lof NaOH added 22.9mL× 1000mL 0.5mol =0.011145molsof NaOH added 0.02229L× 1L −3 Buffer Capacity ofB withaddition ofNaOH =1.32× 10 Calculated valuesfor Buffer SolutionB∧HCl: Volumeof HCl added: 4.96mL InitialpH:4.22 FinalpH:1.96 Molesof NaOH added: 1L =0.00496Lof HCl added 4.96mL× 1000mL 0.5mol =0.00248molsof HCl added 0.00496L× 1L Buffer Capacityof B withrespect¿ additionof HCl= molesof HCladded change ∈ pH ( 0.00248molsof HCladded ) BufferCapacity= ( 4.22−1.96 ) ( ( ) ) ( ) ) ( ( ) −3 Buffer Capacityof A withrespect ¿ additionof NaOH=1.10×10 3. Is there a difference between the buffer capacity of buffers A and B. Explain There is a difference between the buffer capacity of buffers A and B. Buffer A has a smaller capacity value than B. This is evident in the strong acid and base trials added to both of the buffer solutions. 4. Is there a difference in buffer A’s ability to buffer additional acid as compared to its ability to buffer additional base? Explain. Yes, there is a difference as Buffer A is able to buffer additional base more than its ability to buffer additional acid. This is due to the fact that the initial pH of buffer A starts less than 7 = acidic solution. This allows for base to be added with respect to acid added in solution. 5. Answer the following questions as part of your discussion: 1. Say, for example, that you had prepared a Buffer C, in which you mixed 8.203g of sodium acetate, NaC2H3O2, with 100.0 mL of 1.0 M acetic acid. a. What would be the initial pH of Buffer C? pKa=4.76 pH=4.76+log ] ( [[ base acid ] ) pH=4.76+log ] ( [[ 1.0M 0.1M ] ) Therefore,the pH of bufferC=5.76 b. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer’s pH would change less? Calculate the pH after the addition of the base to both buffers B and C. By adding 0.5 M NaOH the buffer solution with a lower pH and higher capacity would have the greater change in pH. The initial pH of buffer C is greater than the initial pH of buffer B, therefore, buffer C would change less in pH as it has less ions to react with the addition of NaOH. pH of Buffer Bafter theaddition of 5.0mLof 5.0M NaOH: −¿ added ¿ 0.5mol =0.0025molOH ( 0.005L ) L ( 0.02L ) 0.1mol =0.002molNaC 2 H 3 O 2 L ( ( ) ) Ice Table values: HA + I C Final NaOH → NaA + H20 0.002 0.0025 0 -0.002 -0.002 +0.002 0 0.0005 0.002 - pH=4.76+log ( 0.0005 0.002 ) pH=4.16 pH of Buffer C after addition of 5.0mL of NaOH: −¿ added 0.5mol ¿ =0.0025molOH ( 0.005L ) L ( 0.02L ) 0.1mol =0.002mols L ( ) ( ) Ice Table values: HA + I C Final NaOH → NaA + H20 0 +0.002 0.002 - 0.002 0.0025 -0.002 -0.002 0 0.0005 ( pH=5.57 + log 0.0005 0.002 ) pH=5.16 2. If you wanted to carry out an experiment at ‘physiological pH’ (7.4) could you suggest an appropriate buffer system? (Hint: The blood buffer system is incredibly complex, and does not make for a good buffer system in the lab) In order to have a good buffer for a physiological pH, it would need to be within +/- 1 of the pKa value from the Ka of blood and the physical pH of 7.4. This would allow for the studying of the buffer system experiment without having to use the incredibly complex buffer (in this example, blood). 3. Why were the buffers asymmetric – that is they seemed to handle added sodium hydroxide with relatively small changes in the pH while small quantities of HCl caused the pH to decrease substantially? The asymmetric buffers meant that the difference in changes of pH with a base added to solution and the addition of an acid were able to be observed. The initial pH of the buffer being less than 7.0, slightly acidic, proved that it takes a significantly smaller amount of acid than base to fill a buffer’s capacity. Thus, the NaOH solution and buffer solution B rose steadily, not showing large quantities of change over an extended period of time. Submit your worksheet on time and to your TA in the dropbox on D2L.