P h and buffer

pH and Buffer
Presented 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 • 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 of
a. Calculation of [H-] chemical applications. In
b. Calculating the base 10 biochemistry, one good
example of buffer solution
log of H-
found in nature is blood
c. pH is negative of the which is present in all
value found in base 10 living organisms.
log

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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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
= 0.1 mol x 136.07 g/mol
KH2PO4 = 13.60 g
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DATA AND RESULTS
SOLUTION C (3rd solution) - 100mL of 1M CH3COOH
Vol. in liter= 0.1 ; M= 1 M ; Mwt. (CH3COOH)= 60 g/mol
= 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 = 5.72ml
SOLUTION D (4th solution)-100mL of 1 M CH3COOHNa+
Vol. in liter= 0.1 ; M= 1 M ; Mwt. (CH3COOHNa+)= 83 g/mol
= 0.1 mol x 83 g/mol
CH3COOHNa+ =8.30 g next

DATA AND RESULTS
SOLUTIONS pH meter pH paper Theoretical pH
1st solution (10 ml distilled 7.22 7.00 7.00
water)
2nd solution (10 ml 3.20 3.00 4.30
distilled water + 1 drop of
HCl)
3rd solution ( 10 ml 7.47 7.00 7.14
phosphate buffer- KH2PO4
and K2HPO4 )

3rd solution ( 10 ml 7.43 7.00 7.14
phosphate buffer- KH2PO4
and K2HPO4 + 1 drop of
HCl)

4th solution ( 10 ml 5.98 6.00 4.74
acetate buffer- CH3COOH
and CH3COOHNa+. )

4th solution ( 10 ml 5.96 5.00 4.74
acetate buffer- CH3COOH
and CH3COOHNa+ + 1
drop of HCl )
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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.
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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.
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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 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

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)
?= 0.01/0.11
M=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. next

Presented by: Aguilar, Princess Alen
Bumagat, Giane Carla
Luis, Ana Patricia
Villanueva, Christian

P h and buffer

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