pH and BufferPresented by: Aguilar, Princess Alen

Bumagat, Giane Carla Luis, Ana Patricia Villanueva, Christian

pH and Buffer

•INTRODUCTION•OBJECTIVES

•METHODOLOGY•DATA AND RESULTS

•DISCUSSION•ANSWER TO THE QUESTION

•CONCLUSION

INTRODUCTION• pH- introduced in

1909 by Sorensen. - it is defined as negative log of Hydrogen ion concentration.

• There are solutions in calculating the pH. Here are some of them:

a. Calculation of [H-]b. Calculating the base

10 log of H- c. pH is negative of the

value found in base 10 log

• Buffer- it is an aqueous solution consisting of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It can resists pH change. Basically, it is use in keeping the pH at a nearly constant value in a wide variety of chemical applications. In biochemistry, one good example of buffer solution found in nature is blood which is present in all

living organisms.

OBJECTIVES

•To illustrate the buffering properties of phosphates and acetates

•To provide the students a sense of how buffers work.

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.

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METHODOLOGY

Get the pH of all four solutions using both the pH meter and pH paper.

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METHODOLOGY

DATA AND RESULTS•PREPARATION OF BUFFERSSOLUTION 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

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DATA AND RESULTSSOLUTION C (3rd solution) - 100mL of 1M CH3COOH

Vol. in liter= 0.1 ; M= 1 M ; Mwt. (CH3COOH)= 60 g/mol

Gram(cpd.)= mol x molar weight of the compund (cpd.)

= 0.1 mol x 60 g/mol CH3COOH =6.00g

To 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/mol

Gram(cpd.)= mol x molar weight of the compund (cpd.)

= 0.1 mol x 83 g/molCH3COOHNa+ =8.30 g

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DATA AND RESULTSSOLUTIONS pH meter pH paper Theoretical pH

1st solution (10 ml distilled water)

7.22 7.00 7.00

2nd solution (10 ml distilled water + 1 drop of HCl)

3.20 3.00 4.30

3rd solution ( 10 ml phosphate buffer- KH2PO4 and K2HPO4 )

7.47 7.00 7.14

3rd solution ( 10 ml phosphate buffer- KH2PO4 and K2HPO4 + 1 drop of HCl)

7.43 7.00 7.14

4th solution ( 10 ml acetate buffer- CH3COOH and CH3COOHNa+. )

5.98 6.00 4.74

4th solution ( 10 ml acetate buffer- CH3COOH and CH3COOHNa+ + 1 drop of HCl )

5.96 5.00 4.74

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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

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

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

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.

ANSWER TO THE QUESTION1. Show the equilibrium for

the ionization of acetic acid and KH2PO4.

*CH3COOH (aq) + H2O (l) -----> H3O+ (aq) + C2H3O2- (aq)

* KH2PO4 -------> K+ + H2PO4

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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]

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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)?= 0.01/0.11M=0.09 CH3COOHNa+

pH = pka + log [salt]/[acid]=4.74 + log [0.09]/[0.009]=5.74

4. 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.

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.

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THE END!Thank You for Listening!

Presented by: Aguilar, Princess Alen

Bumagat, Giane Carla

Luis, Ana Patricia Villanueva,

Christian