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UNIVERSITI KEBANGSAAN MALAYSIA

LABORATORY REPORT
PHYSICOCHEMICAL PROPERTIES OF DRUGS
(NFNF 1213)
TITLE : BUFFERS AND BUFFER CAPACITY
GROUP  : 20
NAME  : NURUL NAZIRA BINTI HUSIN A169005
LECTURER’S NAME : DR LAM KOK WAI
SEMESTER I, SESSION 2018-2019
FACULTY OF PHARMACY
TITLE : BUFFERS AND BUFFER CAPACITY
OBJECTIVE :
To determine the pH of buffer solutions.

To measure the buffer capacity.

To know the correct techniques in handling of pH meter.

DATE : 16 October 2018
INTRODUCTION :
Buffers are compounds that resist changes in pH when a small amount of acid or base is added. Buffer solution is a solution that resists change in pH when a small amount of acid or base is added. Buffer solutions contain a weak acid and its conjugate base (salt) or a weak base and its conjugate acid. For buffer solution that contains weak acid and conjugate base, when a base is added to the solution it reacts with the weak acid to form more conjugate base. When an acid is added to the same solution, it reacts with conjugate base to form more weak acid. In both cases, the pH of solution will remain approximately constant since the number of hydroxide ions, OH- or hydrogen ions, H+ ions in solution is approximately constant.
17526003702050The pH of buffer solutions can be determined using Henderson-Hasselbalch equation which is :
pH = pKa + log A-HAwherepKa: negative logarithm of the dissociation constant for the weak acidA-: substance concentration of the salt (conjugated base)
HA: substance concentration of the weak acid
Buffer capacity (?) is the amount of a strong acid or a strong base that has to be added to 1 litre of a buffer to cause pH change of 1.0 pH unit. The buffer capacity is highest when there are an equal amount of weak acid and conjugate base or weak base and conjugate acid.

5715038100? = ?A /?B?pH where ?A /?B is number of mole of acid or base added and ?pH is changed
of pH observed

EXPERIMENTAL METHOD :
Apparatus :
Vial 20 mL, funnel, pH meter, measurement cylinder 10 mL, volumetric flask 100 mL
Chemicals :
NaH2PO4 0.1 M, Na2HPO4 0.1 M, NaCl 0.1 M, HCl 0.1 M, NaOH standard solution 0.1 M
Experimental procedures :
Directions for use of the pH meter
The pH meter has been already switched on, checked and calibrated by technicians using standard buffer solutions pH 7.0 and 4.0. Any settings cannot be changed.

pH measurement :
The electrode is resting in buffer of pH 7.00 (the green solution). For measurement of pH, the electrode is pulled out from the buffer pH 7.0, rinsed it with distilled water, wiped gently with a piece of tissue paper (just touch, avoid rubbing the glass bulb that would charge it with static electricity) and placed it into the sample solution in plastic container.
Waited until stable reading is achieved (the usual response time of glass electrode is about one minute). The value is written down.
The electrode is pulled out, rinsed again with distilled water, wiped with a piece of tissue paper and placed it into another sample or when the experiment is completed into the buffer pH 7.0.

Preparation prepare 0.1 M of NaCl solution
Students are required to calculate and prepared 0.1 M of NaCl solution in a 100 mL volumetric flask.
Measurement of Buffer capacity
The initial mixtures 1-4 is prepared as shown in the following table :

Vial No 1 2 3 4
Na2HPO4 0.1 M 5 1 9 –
NaH2PO4 0.1 M 5 9 1 –
NaCl 0.1 M – – – 10
Then, the pH of each mixture solution is measured and recorded.

Each vial is added with 1 mL HCl 0.1 M above and mixed it. Then, the pH measured is recorded.
Step (1) to (2) is repeated again by adding with 1 mL NaOH 0.1 M and mixed it. The pH measured is recorded.
RESULTS AND CALCULATIONS :
All the results recorded :
Vial No 1 2 3 4
Na2HPO4 0.1 M (mL) 5 1 9 –
NaH2PO4 0.1 M (mL) 5 9 1 –
NaCl 0.1 M (mL) – – – 10
Acid/conjugate base ratio (mL) 1 : 1 1:9 9:1 –
pH measured 6.26 5.53 7.45 6.34
pH calculated 7.21 6.26 8.16 7.00
Add to all vial with
HCl 0.1 M (mL) 1 1 1 1
pH measured 6.53 6.01 6.66 2.20
pH calculated 7.03 5.98 7.81 2.04
NaOH 0.1 M (mL) 1 1 1 1
pH measured 7.01 6.51 8.86 11.77
pH calculated 7.39 6.61 8.23 11.96
Before addition of HCl/NaOH :
Vial 1 :Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(5) =M2(10) (0.1)(5) =M2(10)
M2 = 0.05 M M2 = 0.05 M
230504917653000
pH = pKa + log saltacid = 7.21 + log 0.050.05 pH = 7.21
Vial 2 : Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(1) =M2(10) (0.1)(9) =M2(10)
M2 = 0.01 M M2 = 0.09 M
222885017780000
pH = pKa + log saltacid = 7.21 + log 0.010.09 pH = 6.26
Vial 3 : Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(9) =M2(10) (0.1)(1) =M2(10)
M2 = 0.09 M M2 = 0.01 M
220027517843500
pH = pKa + log saltacid = 7.21 + log 0.090.01 pH = 8.16
Vial 4 : The NaCl solution is neutral which is pH = 7. It is because sodium ion, Na+ does not
ionize to form hydrogen ion, H+ while chloride ion, Cl- does not ionize to form
hydroxide ion, OH-.

After addition of 1 mL of 0.1 M HCl :
Mole of HCl, n = 0.1 M × 0.001 L
= 0.0001 mol
80962517272000
Vial 1 :pH = pKa+ log saltacid
= 7.21 + log 0.0005 – 0.00010.0005 + 0.0001 pH = 7.03
80962520891500
Vial 2 :pH = pKa+ log saltacid = 7.21 + log 0.0001 – 0.00010.0009 + 0.0001 = math error
The following equation are used as vial 2 cannot use Henderson-Haselbalch equation to determine its pH :
H2PO4 – + H2O ? HPO4 2- + H3O+020383500
Ka = HPO4 2- H3O+H2PO4 –
Ka = x(x)c -x= x2c Where c is concentration,
c = 0.2 mol11 mL ÷ 1000
= 1.818 × 10-5 M
pKa is negative log Ka, thus
pKa = – log Ka7.21 = – log KaKa = 6.166 × 10-8Hence,
6.166 × 10-8 = x21.818 × 10-5x = H3O+ = 1.059 × 10-6 pH = – log H3O+ = – log 1.059 × 10-6 pH = 5.98
81915018097500
Vial 3 :pH = pKa+ log saltacid = 7.21 + log 0.0009 – 0.00010.0001 + 0.0001 pH = 7.81
Vial 4 : NaCl is a neutral solution, so we use
M1V1=M2V2 (0.1)(0.001) = M2 (0.011)
M2 = 9.091 ×10-3 M
pH = – log H3O+ = – log 9.091 ×10-3 pH = 2.04
After addition of 1 mL of 0.1 M NaOH :
Mole of NaOH, n = 0.1 M × 0.001 = 0.0001 mol
81915018161000
Vial 1 :pH = pKa+ log saltacid = 7.21 + log 0.0005 + 0.00010.0005 – 0.0001 pH = 7.39
8286751333500Vial 2 :pH = pKa + log saltacid = 7.21 + log 0.0001 + 0.00010.0009 – 0.0001
pH = 6.61
80010016192400
Vial 3 :pH = pKa + log saltacid = 7.21 + log 0.0009 + 0.00010.0001 – 0.0001 = math error
The following equation are used as vial 3 cannot use Henderson-Hasselbalch equation to determine its pH :
H2PO4 2- + H2O ? H2PO4 – + OH-13335016002000
Kb = H2PO4 – OH-H2PO4 2- Kb = x(x)c -x = x2c Kw= H+OH- = 1.0 × 10-14 Kw= Ka Kb
1.0 × 10-14 = 6.166 × 10-8 Kb
Kb = 1.622 × 10-71.622 × 10-7 = x21.818 × 10-5x = OH- = 1.717 × 10-6 pOH = – log OH- pH + pOH = 14
= – log 1.717 × 10-6pH + 5.77 = 14
= 5.77 pH = 8.23
Vial 4 :
NaCl is a neutral solution, so we use :
M1V1=M2V2 (0.1)(0.001) = M2 (0.011)
M2 = 9.091 ×10-3 M
pOH = – log OH-pH + pOH = 14
= – log 9.091 ×10-3pH + 2.04 = 14
= 2.04 pH = 11.96

QUESTIONS :
By using the Henderson-Hasselbalch equation, the pH values of prepared buffers (pKa = 7.21) are
Before addition of HCl/NaOH :
Vial 1 :Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(5) =M2(10) (0.1)(5) =M2(10)
M2 = 0.05 M M2 = 0.05 M
230504917653000
pH = pKa + log saltacid = 7.21 + log 0.050.05 pH = 7.21
Vial 2 : Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(1) =M2(10) (0.1)(9) =M2(10)
M2 = 0.01 M M2 = 0.09 M
222885017780000
pH = pKa + log saltacid = 7.21 + log 0.010.09 pH = 6.26
Vial 3 : Na2HPO4 (salt) NaH2PO4 (acid)
M1V1 = M2V2 M1V1 = M2V2 (0.1)(9) =M2(10) (0.1)(1) =M2(10)
M2 = 0.09 M M2 = 0.01 M
220027517843500
pH = pKa + log saltacid = 7.21 + log 0.090.01 pH = 8.16
Vial 4 : The NaCl solution is neutral which is pH = 7. It is because sodium ion, Na+
does not ionize to form hydrogen ion, H+ while chloride ion, Cl- does not
ionize to form hydroxide ion, OH-.

After addition of 1 mL of 0.1 M HCl :
Mole of HCl, n = 0.1 M × 0.001 L
= 0.0001 mol
80962517272000
Vial 1 :pH = pKa+ log saltacid
= 7.21 + log 0.0005 – 0.00010.0005 + 0.0001 pH = 7.03
80962520891500
Vial 2 :pH = pKa+ log saltacid = 7.21 + log 0.0001 – 0.00010.0009 + 0.0001 = math error
The following equation are used as vial 2 cannot use Henderson-Haselbalch equation to determine its pH :
H2PO4 – + H2O ? HPO4 2- + H3O+020383500
Ka = HPO4 2- H3O+H2PO4 –
Ka = x(x)c -x= x2c Where c is concentration,
c = 0.2 mol11 mL ÷ 1000
= 1.818 × 10-5 M
pKa is negative log Ka, thus
pKa = – log Ka7.21 = – log KaKa = 6.166 × 10-8Hence,
6.166 × 10-8 = x21.818 × 10-5x = H3O+ = 1.059 × 10-6 pH = – log H3O+ = – log 1.059 × 10-6 pH = 5.98
1304925-2286000Vial 3 :pH = pKa+ log saltacid = 7.21 + log 0.0009 – 0.00010.0001 + 0.0001 pH = 7.81
Vial 4 : NaCl is a neutral solution, so we use
M1V1=M2V2 (0.1)(0.001) = M2 (0.011)
M2 = 9.091 ×10-3 M
pH = – log H3O+ = – log 9.091 ×10-3 pH = 2.04
The pH results of the initial mixtures from the experiment different than the calculated ones is might be due to some factor. Firstly, parallex error. The level of the eye is not perpendicular to the reading scale of the measuring cylinder while pouring solutions. Next, the exact concentration of NaH2PO4 and Na2HPO4 used in the experiment might change when it has been transferred from one apparatus to another. Besides, pH is the negative logarithm of concentration of H+or H3O+. The value is taken from the concentration of the acid used. However, the concentration of the H+ came from both acid and water molecules, H2O ? H++OH-. Thus, when the pH is measured by using a pH meter, the concentration of H+ from water molecules is also measured.

Based on the data from the experiment, the experimental pH values after the addition of HCl are different with the calculated values. This is caused by several factors. Firstly, the apparatus that being used in the experiment is not rinsed before used it. So, the impurities in the apparatus might be mixed with buffer solutions and affect the experimental pH values. Next, the exact concentration and volume of solution used in preparing the solution might be not accurate as calculated due to the possible losing of solution during the transfer of solution. It is possible that some of the buffer solutions were lost as it was transferred from the beaker to the vial by using pipette. Last but not least, the calculated pH values was obtained by using Henderson Hasselbalch equation. This equation is a valid way to predict the pH of a solution regardless of the buffer concentration. However, in reality, this equation is only valid for very dilute solutions. So, this factors might also contribute to the difference of experimental and calculated pH values.

Vial 1 2 3 4
Buffer capacity ? = 0.00010.18= 5.56 ×10-4? = 0.00010.28 =3.57 ×10-4? = 0.00010.35 =2.86 ×10-4? = 0.00014.96 =2.02 ×10-5Vial 1 : ? = Number of mole of acid or base addedChange in pH observed = (0.1)(0.001)(7.21 – 7.03)
= 5.56 ×10-4Vial 2 : ? = Number of mole of acid or base addedChange in pH observed = (0.1)(0.001)(6.26 – 5.98)
= 3.57 ×10-4Vial 3 : ? = Number of mole of acid or base addedChange in pH observed = (0.1)(0.001)(8.16 – 7.81)
= 2.86 ×10-4Vial 4 : ? = Number of mole of acid or base addedChange in pH observed = (0.1)(0.001)(7.00 – 2.04)
= 2.02 ×10-5Based on the calculation of the buffer capacity above, not all examined solutions behave as buffers. Solution in vial 1 had the highest buffer capacity. It is because the solution in vial 1 have equal amounts of weak acid which is NaH2HPO4 and conjugate base which is Na2HPO4. Buffer solutions are considered to be effective when the ratio of A- to HA ranges anywhere between 10:1 and 1:10. So, solution in vial 2 and 3 are considered as effective buffer solution because the ratio of Na2HPO4 to NaH2PO4 is still in the ranges where vial 2 with ratio 1 : 9 while vial 3 with ratio 9 :1. Furthermore, solution in vial 4 had the lowest buffer capacity because it has a drastic change of pH. Thus, solution in vial 4 is not buffer solution since it is a salt solution and does not contain weak acid and conjugated base.

DISCUSSION :
Buffers system based on this experiment are composed of NaH2PO4 that acts as weak acid and Na2HPO4 that is its salt or known as conjugated base. Buffer solutions prepared by mixing 0.1 M NaH2PO4 with 0.1 M Na2HPO4 and then by using pH meter, the pH of buffer solutions was measured. There were 4 vial prepared during this experiment. For vial 1, 5 mL of Na2HPO4 was mixed with 5 mL NaH2PO4 and pH measured was 6.26. For vial 2, 1 mL of Na2HPO4 and 9 mL NaH2PO4 were mixed and pH measured was 5.53. For vial 3, 9 mL of Na2HPO4 was mixed with 1 mL NaH2PO4 and reading on pH meter for this buffer solutions was 7.45. For vial 4, not buffer solutions was prepared but NaCl solutions was prepared. It is prepared by dilution process and we got the pH was equal to 6.34. Then, pH calculated for each vial was obtained by using Henderson-Hasselbalch equation :
pH = pKa + log A-HANext, each vial was added with 1 mL of 0.1 M HCl and followed by addition of 1 mL of 0.1 M NaOH. The pH measured and calculated were obtained by using the same ways as the buffer solutions that contains NaH2PO4 and Na2HPO4. The purpose of addition of NaCl and NaOH to test the buffer solutions whether it is considered as best buffer solution or otherwise. Refer to the definition, best buffer solutions is when it can resists change in pH when a small amount of acid or base is added. When a strong base is added, the acid present in the buffer neutralizes the hydroxide ions (OH-). When a strong acid is added, the base present in the buffer neutralizes the hydrogen ions (H+) or hydronium ions (H3O+). In addition, the best buffers solution have equal amounts of weak acid and conjugate base or weak base and conjugate acid. From this experiment, solutions in vial 1 is the best buffer solutions since the acid/conjugate base ratio was 1 : 1.
We can also know the best buffer solutions by calculate the buffer capacity of the solutions. Buffer capacity is also defined as measure of the efficiency of a buffer in resisting changes in pH. According to the experiment, the buffer capacity for vial 1 was 5.56 ×10-4, for vial 2 was 3.57 ×10-4, for vial 3 was 2.86 ×10-4 and for vial 4 was 2.02 ×10-5. From this data, vial 1 had the highest buffer capacity and vial 4 was the lowest buffer capacity since the difference between pH is too large. Thus, NaCl is not considered as buffer solutions because it does not have weak acid and conjugate base.

Buffer solutions play an important role in biological system. For example, blood as buffer solution. Blood itself tends to be a buffer solution by keeping its pH value constant. Buffer solutions help in the adjustment of the nature of blood. If the alkaline nature of blood increases, buffer solutions tend to decrease the pH value of blood. Hence, when the acidic nature of blood decreases, buffer solutions tend to increase the pH value of blood. If the pH value of blood remains in either alkaline or acidic form then it could lead harmful to a human being. This is could also interrupt the working of some organs.
CONCLUSION :
From the experiment, we can determined the pH of buffer solutions. It can be determined by using pH meter and Henderson-Hasselbalch equation. Indeed, we can actually know the correct techniques in handling of pH meter. There are differences between pH measured and calculated but it already been explained in question section. Next, we can also measured the buffer capacity for each solutions and as the results, solutions in vial 1 had the highest buffer capacity. Last but not least, it is important for us to know how the buffers solutions work because without buffer solutions, our body may undergo a lot of changes.

REFERENCES :
Khan Academy.2015.9 December.Chemistry of buffers and buffers in our blood
https://www.khanacademy.org/test-prep/mcat/chemical-processes/acid-base-equilibria/a/chemistry-of-buffers-and-buffers-in-blood 1 November 2018
Go Life Science.2016.29 September.Buffers: What are the Importance in Biological system? https://www.golifescience.com/buffers-importance/ 3 November 2018
Lumen.2017.11 September. Boundless Chemistry https://courses.lumenlearning.com/boundless-chemistry/chapter/buffer-solutions/
30 October 2018
The Pharmaceutics and Compounding Laboratory.2009.11 April. Buffers and Buffer Capacity. https://pharmlabs.unc.edu/labs/ophthalmics/buffers.htm 1 November 2018

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