Laboratory Report for Butter

OBJECTIVES The objective of the experiment conducted was to give students the understanding the derivation of the Henderson-Hasselbalch from the Ka equation, to be able to use this equation to determine the amount of an acid and conjugate base needed to make a buffer solution at a certain specific pH, and to investigate the relationship between the buffer capacity and concentration. INTRODUCTION Buffer solutions are chemical solutions that resist changes in pH when added with an acid or a base. It is a mixture of a weak acid and its conjugate base. The acidic component of the solution reacts with the added basic substance and the conjugate base reacts with the added acidic substance keeping the pH of the solution stable. This chemical reaction between the buffer components is illustrated by the equilibrium chemical equation below: HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) This equilibrium will be shifted to the left when an acidic compound is added to the buffer solution, and, the equation is shifted to the right when a basic compound is added to the buffer solution. The equilibrium constant expression for the buffer is: Ka=¿ ¿ By rearranging the equation to solve for [H3O+] and taking the negative logarithm of both sides of the equation, the resulting expression will give us the Henderson-Hasselbalch equation: pH = pKa + log ¿ ¿ This equation is the most convenient way to calculate for the pH of the buffer solution, including the amount of acid and conjugate base components of the buffer to get a specific pH value. CALCULATIONS AND DISCUSSIONS Calculation of buffer components The amount of acid and conjugate base components of the buffer is determined using the calculation below. 100 mL phosphate buffer solution with 0.10 M concentration at pH 8 and pH 12.3 were prepared. PART A: pH 8 buffer solution components (Q1. Which compounds did you use for the acid and the base of the buffer? Explain why these two were chosen.) Answer: The rule of thumb for the determination of acid and base component of the buffer solution is to use dissociation constant with pKa value near the pH value of the buffer you want to prepare. For pH 8 phosphate buffer, the pKa used is 7.20 which is from negative log of the Ka2 = 6.3×10-8 which implies that the second proton of H3PO4 is dissociated to get HPO42-. Hence, based on the available compounds in the laboratory, the NaH2PO4∙H2O was selected as the acid component and the Na2HPO4∙7H2O as the base component. The equilibrium equation for the buffer is: H2PO4- + H2O ⇌ HPO42- + H3O+ Based on the above equilibrium equation, the compounds used for this buffer are NaH2PO4∙H2O as the acidic component and Na2HPO4∙7H2O as the conjugate base component. Using Henderson-Hasselbalch equation, pH = pKa + log ([HPO HPO42-]/[H2PO4-]) the amounts of acid and conjugate base components of the buffer was calculated. 8 = 7.2 + log ([HPO4 2-]/[H2PO4-]) 0.8 = log ([HPO42-]/[H2PO4-]) 100.8 = ([HPO42-]/[H2PO4-]) 6.31 = ([HPO42-]/[H2PO4-]) 6.31[H2PO4-] = [HPO42-] (Q2-3 Show the calculations for the acid and base components’ concentration, number of moles, and mass) a. Calculation of concentrations of acid and basic component Concentration of NaH2PO4∙H2O(acidic component) Since [H2PO4-] + [HPO4 2-] = 0.10 M [H2PO4-] + 6.31[H2PO4-] = 0.10 M [H2PO4-] (7.31) = 0.10 M  0.0137 M Concentration of Na2HPO4∙7H2O (basic component) [H2PO4-] + [HPO42-] = 0.10 M (0.0137 M) + [HPO42-] = 0.10 M [HPO42-] = 0.0863 M b. Calculation of number of moles and mass b.1 Acidic component NaH2PO4∙H2O Number of moles of NaH2PO4∙H2O Moles of H2PO4- = (0.0137 mol/L) x 0.1 L  0.00137 moles NaH2PO4∙H2O Mass of NaH2PO4∙H2O Mass of H2PO4- = 0.00137 moles x (137.99 g/mol) = 0.189 g NaH2PO4∙H2O b.2 Basic component Na2HPO4∙7H2O Number of moles of Na2HPO4∙7H2O Moles of HPO42- = (0.0863 mol/L) x 0.1 L  0.00863 moles Na2HPO4∙7H2O Mass of Na2HPO4∙7H2O Mass of HPO42- = 0.00863 moles x (268.07 g/mol) = 2.31 g Na2HPO4∙7H2O Q4 Table 1: Summary of concentration, number of moles, and mass calculations of acidic and basic components of pH 8 phosphate buffer solution Number of moles, Concentration, Mass, g Component mol mol/L NaH2PO4∙H2O 0.0137 0.00137 0.189 Na2HPO4∙7H2O 0.0863 0.00863 2.31 Table 2: Buffer pH adjustment and buffer capacity # of drops of NaOH pH after Solution Initial pH to adjust adjustment the pH pH 8 buffer 7.87 4 8.01 Deionized water pH after dilution pH with 3 drops of added acid pH with 3 drops of added base 8.02 7.99 8.04 3.01 10.05 (Q5 Would you expect this buffer to be better at resisting a change in pH upon addition of acid upon addition of base, or about the same for both? Why?) The buffer capacity of the pH 8 phosphate buffer solution can be observed from the data tabulated in table 2. Upon addition of 3 drops of acid, the pH of the buffer decreased by 0.03 units from 8.02 to 7.99 pH value. Upon addition of 3 drops of base, the pH of the buffer increased by 0.02 units from 8.02 to 8.04 pH value. Hence, this buffer is more resistant to the addition of base. PART B: pH 12.3 buffer solution components (Q6 Which compounds did you use for the acid and the base of the buffer? Explain why these two were chosen.) Answer: The rule of thumb for the determination of acid and base component of the buffer solution is to use dissociation constant with pKa value near the pH value of the buffer you want to prepare. For pH 12.3 phosphate buffer, the pKa used is 12.38 which is from negative log of Ka3 = 4.2×10-13 which implies that the third proton of H3PO4 is dissociated to get PO42-. Hence, based on the available compounds in the laboratory, the Na2HPO4∙7H2O was selected as the acid component and the Na3PO4∙12H2O as the base component. The equilibrium equation for the buffer is: HPO4- + H2O ⇌ PO42- + H3O+ Using Henderson-Hasselbalch equation, pH = pKa + log ([PO PO4 2-]/[HPO4-]) the amounts of acid and conjugate base components of the buffer was calculated. 12.3 = 12.38 + log ([PO42-]/[HPO4-]) -0.08 = log ([PO42-]/[HPO4-]) 10-0.8 = ([PO42-]/[HPO4-]) 0.83 = ([PO42-]/[HPO4-]) 0.83[HPO4-] = [PO42-] (Q7-8 Show the calculations for the acid and base components’ concentration, number of moles, and mass) a. Calculation of concentrations of acid and basic component Concentration of Na2HPO4∙7H2O (acidic component) Since [HPO4-] + [PO42-] = 0.10 M [HPO4-] + 0.83[HPO4-] = 0.10 M [HPO4-] (1.83) = 0.10 M  0.0546 M Concentration of Na3PO4∙12H2O (basic component) [HPO4-] + [PO42-] = 0.10 M (0.0546 M) + [PO4 2-] = 0.10 M [HPO42-] = 0.0453 M b. Calculation of number of moles and mass b.1. Acidic component Na2HPO4∙7H2O Number of moles of Na2HPO4∙7H2O Moles of HPO4- = (0.0546 mol/L) x 0.1 L  0.00546 moles Na2HPO4∙7H2O Mass of Na2HPO4∙7H2O Mass of HPO4- = 0.00546 moles x (268.07g/mol) = 1.46 g Na2HPO4∙7H2O b.2 Basic component Na3PO4∙12H2O Number of moles of Na3PO4∙12H2O Moles of PO42- = (0.0453 mol/L) x 0.1 L  0.00453 moles Na3PO4∙12H2O Mass of Na3PO4∙12H2O Mass of PO42- = 0.00453 moles x (380.12 g/mol) = 1.72 g Na3PO4∙12H2O Q9 Table 3: Summary of concentration, number of moles, and mass calculations of acidic and basic components of pH 8 phosphate buffer solution Number of moles, Concentration, Mass, g Component mol mol/L Na2HPO4∙7H2O 0.0546 0.00546 1.46 Na3PO4∙12H2O 0.0453 0.00453 1.72 Table 4: Buffer pH adjustment and buffer capacity # of drops of NaOH pH after Solution Initial pH to adjust adjustment the pH pH 12.3 12.03 15 12.3 buffer Deionized water pH after dilution pH with 3 drops of added acid pH with 3 drops of added base 12.29 12.27 12.3 Q10 The buffer capacity of the pH 12.3 phosphate buffer solution can be observed from the data tabulated in table 2. Upon addition of 3 drops of acid, the pH of the buffer decreased by 0.02 units from 12.29 to 12.27 pH value. Upon addition of 3 drops of base, the pH of the buffer is almost the same from 12.29 to 12.3 pH value. Hence, this buffer is more resistant to the addition of base. PART C: Dilute buffer of pH 8.0 Q11 Show the calculations to determine the volume of the 0.1 M buffer from Part A needed to make 100 mL of a 0.01 M buffer. pH 8 buffer initial concentration = 0.10 M (C1) C1V1 = C2V2 0.10 M x (V1) = 0.01 M x 100 mL, find V1 V1 = (0.01M x 100mL)/0.10 M = 10 mL Hence, you need 10 mL of 0.10 M pH 8.0 buffer solution to make 100 mL of 0.01 M pH 8.0 buffer solution. Q12 Table 5: Buffer capacity of diluted buffer of pH 8.0 Solution Initial pH 0.01 M pH 8.0 buffer 8.11 pH with 3 drops added acid 7.61 pH with 3 drops added base 8.49 In comparison to the 0.10 M buffer pH 8.0 solution, the 0.01 M buffer solution is less resistant to the addition of an acid or a base. Hence, we can conclude that when the concentration of the buffer at any given pH is decreased, the effect of the addition of an acid or base to it is greater as can be seen in the changes in pH value. INDIVIDUAL REPORT PORTION 1. Calculate the expected pH of a strong acid solution (HCl) at concentrations of 0.1 M and 0.01 M. pH = -log [H+], [H+] = concentration of HCl 1.a pH of HCl at 0.1 M pH = -log [0.1M] = 1 Hence, the pH of HCl at 0.1M concentration is pH 1. 1.b pH of HCl at 0.01 M pH = -log [0.01M] = 2 Hence, the pH of HCl at 0.01M concentration is pH 2 2. In part C, the pH 8.0 buffer was diluted from 0.1 M to 0.01 M. Compare the predicted pH change for the strong acid dilution to the observed pH change for the buffer. Buffer capacity is the measurement of the efficiency of the buffer solution to resist pH changes upon addition of an acid or a base. The 0.01 M buffer solution is less resistant to the addition of an acid or a base when compared to the 0.10 M buffer pH 8.0 solution. Hence, we can conclude that when the concentration of the buffer at any given pH is decreased, the effect of the addition of an acid or base to it is greater as can be seen in the changes in pH value. 3. In parts A and B, you adjusted the pH before increasing the volume to 100 mL. Use the Henderson-Hasselbalch equation to explain why you should not expect the buffer pH to be sensitive to being diluted. Then Henderson-Hasselbalch equation is pH = pKa + log ¿ ¿ As seen in the equation above, the pH of the buffer solution is only dependent on the amount of acid and conjugate base components present in the solution. That is why during the pH adjustments, drops of a strong acid or strong base were used. Hence, diluting the buffer solution with water does not affect the pH value of the buffer solution. 4. How did your observation of the diluted buffer in part C compare to this conclusion? Based on the data gathered, there was no significant change in pH value for the diluted pH 8.0 solution from 0.10 M to 0.01 M concentration. Hence, the prediction given from the HendersonHasselbalch equation holds true. 5. Compare the results of the pH sensitivity data in parts A and C. Explain why the buffer capacity depends on the concentration of the buffer. Based on the pH sensitivity in parts A and C, the concentration of the buffer plays an important role to its buffer capacity. As the concentration of the buffer decreases, the buffer capacity also decreases. This is because of the relative amount of the dissociated acid and conjugate base components present that counters the effect of any added acid or base to the solution. The lower the concentration, the lower the acid and conjugate base present and hence, the lower is the pH change resistivity.