P h and buffer

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P h and buffer

  1. 1. pH and BufferPresented by: Aguilar, Princess Alen Bumagat, Giane Carla Luis, Ana Patricia Villanueva, Christian
  2. 2. pH and Buffer • INTRODUCTION • OBJECTIVES • METHODOLOGY • DATA AND RESULTS • DISCUSSION• ANSWER TO THE QUESTION • CONCLUSION
  3. 3. INTRODUCTION• pH- introduced in 1909 • Buffer- it is an aqueous by Sorensen. solution consisting of a - it is defined as mixture of a weak acid and its conjugate base or a weak negative log of Hydrogen base and its conjugate acid. It ion concentration. can resists pH change.• There are solutions in Basically, it is use in keeping calculating the pH. Here the pH at a nearly constant are some of them: value in a wide variety ofa. Calculation of [H-] chemical applications. Inb. Calculating the base 10 biochemistry, one good example of buffer solution log of H- found in nature is bloodc. pH is negative of the which is present in all value found in base 10 living organisms. log
  4. 4. OBJECTIVES• To illustrate the buffering properties of phosphates and acetates• To provide the students a sense of how buffers work.
  5. 5. METHODOLOGY• First, preparation of both the phophate buffer which include the KH2PO4 and K2HPO4 . While the acetate buffer contains the CH3COOH and CH3COOHNa+.• To be able to get the correct amount of the compound needed to be able to prepare a 100 ml of each buffer.The solution used is: gram= mol x molar weight of the compund* Mol is obtained by dividing the molarity with the Volume in liter needed for the preparation of the buffers. next
  6. 6. METHODOLOGY Get the pH of all four solutions using both the pH meter and pH paper. next
  7. 7. METHODOLOGY
  8. 8. DATA AND RESULTS • PREPARATION OF BUFFERS SOLUTION A (1st solution) - 100mL of 1 M K2HPO4 Vol. in liter= 0.1 ; M= 1 M ; Mwt. (K2HPO4)= 174.17 g/mol Gram(cpd.)= mol x molar weight of the compund (cpd.) = 0.1 mol x 174.17 g/mol K2HPO4 = 17.40 g SOLUTION B (2nd solution) -100mL of 1 M KH2PO4 Vol. in liter= 0.1 ; M= 1 M ; Mwt. (KH2PO4 )= 136.07 g/mol Gram(cpd.)= mol x molar weight of the compund (cpd.) = 0.1 mol x 136.07 g/mol KH2PO4 = 13.60 g next
  9. 9. DATA AND RESULTSSOLUTION C (3rd solution) - 100mL of 1M CH3COOHVol. in liter= 0.1 ; M= 1 M ; Mwt. (CH3COOH)= 60 g/molGram(cpd.)= mol x molar weight of the compund (cpd.) = 0.1 mol x 60 g/molCH3COOH =6.00gTo get the Volume: V= M/D Density (D)= 1.048 g/ml =6 g / 1. 048 g/ml V = 5.72mlSOLUTION D (4th solution)-100mL of 1 M CH3COOHNa+Vol. in liter= 0.1 ; M= 1 M ; Mwt. (CH3COOHNa+)= 83 g/molGram(cpd.)= mol x molar weight of the compund (cpd.) = 0.1 mol x 83 g/molCH3COOHNa+ =8.30 g next
  10. 10. DATA AND RESULTS SOLUTIONS pH meter pH paper Theoretical pH1st solution (10 ml distilled 7.22 7.00 7.00water)2nd solution (10 ml 3.20 3.00 4.30distilled water + 1 drop ofHCl)3rd solution ( 10 ml 7.47 7.00 7.14phosphate buffer- KH2PO4and K2HPO4 )3rd solution ( 10 ml 7.43 7.00 7.14phosphate buffer- KH2PO4and K2HPO4 + 1 drop ofHCl)4th solution ( 10 ml 5.98 6.00 4.74acetate buffer- CH3COOHand CH3COOHNa+. )4th solution ( 10 ml 5.96 5.00 4.74acetate buffer- CH3COOHand CH3COOHNa+ + 1drop of HCl ) next
  11. 11. DATA AND RESULTS • To get the pH of water: Kw= [H+] [OH-] pH = - log [ 1E-7] = 7.00 • To get the pH of water after adding 1 drop of HCl. H2O + HCl --------> H3O + Cl 1 drop = 0.00005L [H+]= 1E-7 + 0.00005 pH = -log [ 0.00005] =4.30 • To get the pH of the phosphate buffer- 0.005 L K2HPO4 + 00.005 KH2PO4 Dilution formula: (M1V1)= (M2V2) (1mol)(0.005ml)=(?)(0.01ml) ?mol= 0.005ml/ 0.01ml =0.5M of K2HPO4 and KH2PO4 • pH= pka + log [salt]/ [acid] =7.2E-8 + log [0.5]/[0.5] = 7.14 • To get the pH of the acetate buffer- 0.005 L CH3COOH + 0.005 L CH3COOHNa+ Dilution formula: (M1V1)= (M2V2) (1mol)(0.005ml)=(?)(0.01ml) ?mol= 0.005ml/ 0.01ml =0.5M of K2HPO4 and KH2PO4 • pH= pka + log [salt] / [acid] =1.8E-5 + log [0.5]/[0.5] =4.74
  12. 12. DISCUSSION• In every experiment with calculations, we should have precise and accurate data and values to be able to get a correct and successful results. In this experiment, we need to have the correct values for the volume, molarity of the solution to have the exact preparation of the solution and the buffers- both the phosphate and acetate buffer.• In the experiment, we focus on the effect of buffer solution. In which buffer is said to be a combination of weak acid or weak base and its salt.• The 1st solution, which has the distilled water will be the reference solution since water has the neutral pH of 7.00. Using pH meter, the pH is 7.22 and 7.00 using the pH paper. next
  13. 13. DISCUSSION• The second solution was added with a drop of HCl and as expected the pH of the solution dropped since HCl is known to be a strong acid. The pH meter result is 3.20 and 3.00 in pH paper.• The third solution which contains the 10 ml phosphate buffer (5 ml KH2PO4 and 5 ml K2HPO4 ), we obtain a pH of 7.47 using pH meter and 7.00 using pH paper. We then add 1 drop of HCl in the solution and as expected, the pH drops into 7.43 using the pH meter and 7.00 using the pH paper. This shows that the phosphate buffer is effective since it resist a large change in the pH value and in fact, it is still in the pH range from the theoretical pH which is 7.14. next
  14. 14. DISCUSSION• The fourth solution has the 10 ml acetate buffer (CH3COOH and CH3COOHNa+ ), we obtain a pH of 5.98 using pH meter and 6.00 using pH paper. We then add 1 drop of HCl in the solution and as expected, the pH drops into 5.96 using the pH meter and 5.00 using the pH paper. It shows that the obtained pH is near to the theoretical pH of 4.74 since the maximum pH of it is 5.74 . Hence it is computed by adding or subtracting 1 to the final pH value.• Based on all the results gathered, it shows that it has a small difference from the obtain or experimental value to the theoretical value. Thus, it shows a good result. One factor that can affect the result obtain is the instrumental error since the pH meter available and use in the experiment is defective.• The Henderson-Hassellbach equation also gave a big factor in our computation without the knowledge about this, it is very hard for us to compute and even analyze the resulting pH of a given buffer.
  15. 15. ANSWER TO THE QUESTION 1. Show the equilibrium for the ionization of acetic acid and KH2PO4. *CH3COOH (aq) + H2O (l) -----> H3O+ (aq) + C2H3O2- (aq) * KH2PO4 -------> K+ + H2PO4 - next
  16. 16. 2. Derive Henderson- Hesselbach Equation ▫Ka = [H=][A-] [HA] ▫-log Ka= -log = [H=][A-] [HA] -log Ka= - log [H] –log [A-]/[HA] pKa= pH –log [A]/[HA] pH= pka + log [A-]/ [HA] next
  17. 17. 3. An acetate buffer was prepared by mixing 10 mL of 0.1M acetic acid and 100 mL of 0.1M sodium acetate. What is the pH of the buffer solution.Dilution formula: (M1V1) = (M2V2) (0.1M acetic acid)(0.01mL)=(?)(0.11mL) ?= 0.001/0.11 M=0.009 CH3COOH (0.1M sodium acetate)(0.1mL)=(?)(0.11mL) ?= 0.01/0.11 M=0.09 CH3COOHNa+ pH = pka + log [salt]/[acid] =4.74 + log [0.09]/[0.009] =5.744. Can a buffer solution be prepared from a mixture of NaNo3 and HNO3? Explain. *Technically, from the meaning of buffer it must a combination of weak base/acid and its salt, the combination of NaNo3 and HNO3 is a strong acid and salt mixture thus it will not form a buffer solution.
  18. 18. CONCLUSION• Buffer can be said to be effective if it can maintain the pH of a certain solution from its pH range such that the Phosphate buffer is very good buffer for maintaining the pH of blood which is 7.4 since the maximum pH of the phosphate buffer is 8.4 based from its range thus, buffers like acetate buffer is good for particular solution which has the maximum range of 5.74 and minimum range of 3.74.• Moreover knowing the principle of buffers and the Henderson-Hassellbach equation is very crucial for buffer preparation and for better understanding. next
  19. 19. Presented by: Aguilar, Princess Alen Bumagat, Giane Carla Luis, Ana Patricia Villanueva, Christian

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