1.0 SUMMARY

The experiment was done to prepare buffer solutions and test the effects of added acid or base

upon the pH. Also this experiment is done to review acid-base chemistry and calculation. The

result obtained was according to the theory where the buffer solution shows how it works even

when there are acid and base were added. The buffer solution will contain things which will

remove any hydrogen ions or hydroxide ions that might add to it otherwise the pH will change .

But still the pH value will slightly changes due to the equilibrium. In the experiment also, there

are error occurred. It can be seen in the result where the measured pH is different than the actual

pH. In order to eliminate the error, the pH meter must be calibrated before the experiment started

and the preparation of buffer must be done carefully therefore there are no errors occurred in

measurement. As a conclusion, the objectives of the experiment was achieved and the buffer

solution manage to make sure the pH is not changing so much and always in the acceptable

range.

2.0 METHODOLOGY2.1 Preparation of buffer solution

500 ML of a 0.1 M buffer solution was prepared with pH of 5.0 using acetic acid, HC2H3O2 and sodium acetate,

NaC2H3O2

Volume combine method

0.1M of acetic acid is make up and conjugate 0.1 M of sodium acetate and the volumes of two solution is combined to get the

correct ration since the total molarity must be 0.1M

When using the total volume 500mL, apply Henderson-hasselbalch equation to calculate the ratio.

Firstly,appropriate volume of solution calculated frequently and to became the pH desired,it must add other solution until

correct

When adding the second solution,pH of the solution need to monitor frequently

Do not add water to make up the volume when end up with more or less than 500mL of solution

Mass method are used in this experiment

Mass of the acetic acid is calculated

Mass of the acetate ion that are required to make a pH 5.0 solution that has a combined molarity of 0.1M

To determine the molar ratio of acetic acid to acetate ion,Handerson-Hasselbalch is using.

Moles of acetic acid + moles of acetate ion /0.5L=0.1M

each substance is weight out and add distilled water to make a 0.5L solution

2.2 Testing buffer solution

pH of buffer is measure. (Place 25.0 ml of buffer in a clean 100ml beaker. pH of the solution can be determine by using pH electrode)..

100 mL beakers is dry and add 50 mL of deionized water into each buffer.

pH of each buffer was measure and record.

Add 1.0 mL of 1 M HCL for 1 beaker and add 1.0 mLof 1 M Naoh for other beaker.

Measure the pH of each solution

Repeat the procedure in the above graph using buffer solution instead of the deionized water.

3.0 RESULT

Table 3.1: Result for blank

Solution pHBuffer 4.90 4.90Water 6.90 6.90

Table 3.2: Result for solution and add with 1 ml of HCL

Solution pHBuffer 4.60Water 2.90

Table 3.3: Result for solution and add with 1 ml of NaOH

Solution pHBuffer 5.30Water 11.70

Table 3.4: Comparison between measured and calculated pH

Solution pH measured pH calculated

Buffer + NaOH 5.30 5.03

Buffer + HCl 4.60 4.50

Table 3.5: pH changes in water and buffer solution

Water (add HCl) Buffer ( add HCl) Water (add NaOH) Buffer (add NaOH)

Before add 6.90 4.90 6.90 4.90

After add 2.90 4.60 11.70 5.30

Calculation for pH of buffer after adding acid or base

To calculate pH buffer after adding base (NaOH)

C6H8O7 (aq )+OH−¿ (aq )→H 2O (l )+C6 H 7O7−¿ ( aq ) ¿¿

Since mol OH- = 1 M X 1 mL = 1 mol/L X 0.001 L = 0.001 mol

C6H 8O7 OH−¿¿ C6H 7O7−¿ ¿

Initial 0.0185 mol 0.001 mol 0.0315 mol

Change - 0.001 mol - 0.001 mol + 0.001 mol

Final 0.0175 mol 0 0.0325 mol

C6H 8O7 (aq )⇋H+¿ (aq )+C6 H 7O 7−¿ ( aq ) ¿¿

C6H 8O7 H+¿¿ C6H 7O7−¿ ¿

Initial 0.0175 M 0 0.0325 M

Change - x M + x M + x M

Final (0.0175 – x) M

x M (0.0325 + x) M

Since pKa = 4.77 and pKa = - log (Ka)

pKa=−log (Ka )

Ka=anti−log ( pKa )

Ka=anti−log ( 4.77 )

Ka=1.698×10−5

Ka=¿¿

1.698×10−5=[x ] [ (0.0325+x ) ]

[ (0.0175– x ) ]

(2.97×10−7 )−(1.698×10−5 ) x=0.0325 x+x2

x2+0.0325 x+( 1.698×10−5 ) x−(2.97×10−7 )=0

x=9.135×10−6 M

Since H+ = x

pH=−loq ¿

pH=−loq [9.135×10−6 ]

pH=5.03

To calculate pH buffer after adding acid (HCl)

C6H8O7 (aq )+H+¿ (aq )→ H2 (g )+C6H 7O7−¿ (aq )¿ ¿

Since mol H+ = 1 M X 1 mL = 1 mol/L X 0.001 L = 0.001 mol

C6H 8O7 H+¿¿ C6H 7O7−¿ ¿

Initial 0.0185 mol 0.001 mol 0.0315 mol

Change - 0.001 mol - 0.001 mol + 0.001 mol

Final 0.0175 mol 0 0.0325 mol

C6H8O7 (aq )+H2O (l )⇋H 3O+¿ (aq) +C6H 7O7

−¿ (aq ) ¿ ¿

C6H 8O7 H 3O+¿¿ C6H 7O7

−¿ ¿

Initial 0.0175 M 0 0.0325 M

Change - x M + x M + x M

Final (0.0175 – x) M

x M (0.0325 + x) M

From d, Ka = 1.698×10−5

Ka=¿¿

1.698×10−5=[x ] [ (0.0175– x ) ]

[ (0.0325+x ) ]

(5.5185×10−7 )−(1.698×10−5 ) x=0.0175 x+x2

x2−0.0175 x+(1.698×10−5 )x−( 5.5185×10−7 )=0

x=3.159×10−5M

Since H 3O+¿¿

= x

pH=−log¿

pH=−loq [3.159×10−5 ]

pH=4.5

CALCULATION

A = Conjugate Base

HA = Weak Acid

Given data: pKa = 4.77

pH = 5.0

pH = pKa + log¿¿¿

5.0 = 4.77 + log¿¿¿

0.23 = log¿¿¿

¿¿¿ = 1.698

¿¿ = 1.698 [HA] -------

Given data: molarity = 0.1 M

Volume = 500 ml

[ HA ]+¿¿¿ = molarity

[ HA ]+¿¿¿ = 0.1 M

Substitute ¿¿ with equation

[HA ]+[1.698 HA ]500ml

= 0.1

HA + 1.698 HA = 0.05

2.698 HA = 0.05

HA = 0.0185 moles

Substitute value of HA into

A = 1.698 (0.0185)

= 0.0314 moles

1

1

1

Calculation for mass method

Data given:

Molecular weight of citric acid = 210.14

Molecular weight of sodium citrate = 294.10

Mass of Citric Acid

Mass = Moles x Molecular weight

= 0.0185 x 210.14

= 3.8876 g

Mass of Sodium Citrate = 9.2347 g

Calculation for combining volume method

Molarity = molesvolume

Volume = moles

molarity

Volume of Citric Acid

Volume = 0.0185

0.1

= 0.185 L

Volume of Sodium Citrate = 0.500 – 0.185

= 0.315 L

4.0 DISCUSSION

The experiment was done to prepare buffer solutions and test the effects of added acid or base

upon the pH. Also this experiment is to review acid-base chemistry and calculation. A buffer

solution is one which resists changes in pH when small quantities of an acid or an alkali are

added to it. A buffer solution has to contain things which will remove any hydrogen ions or

hydroxide ions that you might add to it - otherwise the pH will change. Acidic and alkaline

buffer solutions achieve this in different ways. The change in the buffer agent initially causes

little change in the buffer solution’s pH. As the proportion of a buffer agent’s HA and A- changes

due to the addition of other acids and bases, the buffering capacity of the solution decreases.

Eventually, a buffer agent can be used up to the point where it can no longer significantly resist

pH changes, meaning it is no longer useful as a buffer solution. In this experiment, the mass

method was used in order to determine the amount of citric acid and sodium citrate. The mass for

citric acid used is 3.8876 g and the mass of sodium citrate used is 9.2347 g. Then the mass was

added together and dissolve in water in 500 mL volumetric flask. The initial pH reading for

blank buffer solution is 4.90 while for water is 6.90. The pH reading is slightly low than desired

pH. This is happen because of error occurred during the preparation. After that 1ml of acid (HCl)

and 1 ml of base (NaOH) was added to 50 ml of buffer solution and 50 ml of water. After the

addition of acid and base into the solution, the reading of pH was measured. For acid addition,

the pH measured for buffer solution is 4.6 while water is 2.9. For base addition, the pH measured

for buffer solution is 5.30 while for water is 11.70. The actual reading for acid addition in buffer

is 4.50 while for base addition is 5.03. From the result obtained, it can be seen that the measured

pH is not same with the calculated pH. This happened because of pH meter that is not calibrated

and also error in preparation of buffer solution. From this two possible causes will lead to error

in pH reading. The theory also stated that a buffer solution is one which resists changes in pH

when small quantities of an acid or an alkali are added to it. For acidic buffer solution, when acid

was added into it, the buffer solution must remove most of the new hydrogen ions. Since most of

the new hydrogen ions are removed, the pH won't change very much but because of the

equilibrium involved, it will fall a little bit. From this theory, the result obtained is according to

the theory where the pH of the buffer solution drops from 4.90 to 4.60 when the acid was added.

And the change for the pH is not too big and still can be accepted. While when base was added

to buffer solution, the buffer solution must remove most of the new hydroxide ions. When most

of the new hydroxide ions are removed, the pH doesn't increase very much. From the result

obtained for base addition, the pH of the buffer solution was slightly increased from 4.90 to 5.30.

The increasing of the pH is no too much and acceptable. Therefore the result obtained for acid

and base addition into buffer solution is good and followed the theory. Next are the pH changes

for the water. The purpose for observing the changes of pH in water is to compare the result for

acid and base addition in the water with acid and base addition in the buffer solution. From the

result, when the acid was added into the water, the pH for water was decreasing much from 6.90

to 2.90 and when the base was added into the water, the pH water increasing greatly from 6.90 to

11.70. From this value, it can be seen how the buffer solution actually work. Water is not a

buffer solution; therefore it will not remove any ion. When the acid was added, the new

hydrogen ion will not be removed but it will be added to the hydrogen ion from dissociation of

water. And because of that, the pH will drop greatly. Same goes to when base was added to the

water. The new hydroxide ion will not be removed but will add with the hydroxide ion from

water. Therefore, the pH will increase greatly. Truly all the result obtained is following the

theory concept.

TUTORIAL

QUESTION 1

Describe how you would prepare a (CH3CO2H / CH3CO2-) buffer with a pH of 5.2 given that the

pKa (CH3CO2H / CH3CO2-) is 4.76

First step: - Calculating mole ratio of CH3CO2H / CH3CO2

pH=p Ka+ log¿¿

Second step: - finding total mole required

Third step:- calculate the amount of mass needed each from CH3CO2H / CH3CO2

`

2.7542=¿¿

A−¿:HA¿

1 :2.7542

mol1:mol2

5.2=4.76+ log¿¿

0.44=log¿¿

log−10.44=¿¿

I f 3.7542→0.05mole,

Acetic acid ,2.7542→2.75423.7542

×0.05

¿0.03668mol

Sodiumacetate ,1→1

3.7542×0.05

¿0.013318mol

mol1+mol2

0.5 L=0.1mol

1L

mol1+mol2=0.05mol

mass=n×mw

Aceticacid 0.03668mol ×294.10 gmol−1=10.7875g

Sodiumacetate ,0.013318mol×210.14 gmol−1=2.7986g

QUESTION 2

What is buffer capacity? What determines the capacity of a buffer?

Buffer capacity

Buffer capacity can be defined as maximum amount of either strong acid or strong base that can

be added before a significant change in the pH will occur. Buffer capacity can be also defined as

quantity of strong acid or base that must be added to change the pH of one liter of solution by

one pH unit. Buffering capacity refers to water's ability to keep the pH stable as acids or bases

are added. pH and buffering capacity are intertwined with one another although one might think

that adding equal volumes of an acid and neutral water would result in a pH halfway in between,

this rarely happens in practice.

What determines the capacity of a buffer?

Determine the capacity of a buffer to compared to the pH of the solution because the higher the

concentration, the larger the buffering capacity. The buffer is compared to the pH of the solution

when the closer the buffer to pKa the value of the buffer capacity is greater increases.

5.0 POSSIBLE ERRORS AND INACCURACY

During this experiment was done, the common error happened was the measured pH is not same

with the actual pH. There are two possible things that can lead to this error. The uncalibrated pH

meter and also the error happen during the preparation of buffer. For the pH meter, to overcome

this problem, the pH meter should be calibrated first before the experiment was run and the pH

electrode should always clean by rinsing it with deionised water. While from the preparation of

buffer; when the mass or volume of the component to make the buffer solution was measured,

make sure it must be measured correctly. Any slight changes in the mass or volume will result

the changes in pH reading and deviate from the desired pH value. Other thing that can be done to

avoid error is make sure the apparatus is clean and also all the acid and base addition must be

done inside the fume cupboard for safety.

6.0 CONCLUSION

As a conclusion, all the objectives of this experiment were achieved. The buffer solution show

how it work when there are acid and base were added into it; where the pH of the buffer is

changing just a little bit when there are acid or base added into it. And the result obtained is

following the theory where a buffer solution is one which resists changes in pH when small

quantities of an acid or an alkali are added to it also a buffer solution has to contain things which

will remove any hydrogen ions or hydroxide ions that you might add to it otherwise the pH will

change. But still due to equilibrium the pH will lightly drop or increase. The experiment was

done successfully.

7.0 REFERENCES

1. Buffer, TutorVista.com, [Online]. [Accessed 5th September 2011]. Available from

http://www.tutorvista.com/content/chemistry/chemistry-iii/ionic-equilibrium/buffers.php

2. Buffer Solution, New World Encyclopedia, [Online]. [Accessed 5th September 2011].

Available from http://www.newworldencyclopedia.org/entry/Buffer_solution

3. What are buffer solutions used in? eHow, [Online]. [Accessed 5th September 2011].

Available from http://www.ehow.com/how-does_4912029_what-buffer-solutions-

used.html

4. Buffer Solution – Introduction and Uses