1. To understand the acid-base chemistry.
2. To prepare and evaluate a buffer system
3. To measure the buffering capacity of two types of isotonic drinks. Introduction:
There are acid-forming, basic forming and neutral food, however the acid or alkaline properties of a food is unable to judge by the actual acidity of the food itself. For example, citrus fruits such as lemon are acidic, but they are alkaline-forming when we consume and digest it. Therefore the food acidity or alkalinity only can determined when it is break down and digested. To maintain a healthy diet, acid-base balance diet is very important.
An acid-base buffer is a solution that lessens the impact on pH from addition of acid and base. The essential feature of a buffer means that it consists of high concentrations of the acidic (HA) and basic (A-) components. When small amount of H3O+ or OH- ions are added to the buffer, they cause a small amount of one buffer component to convert into the other, which change the relative concentration of the two component. For any weak acid, HA, the dissociation equation and Ka expression are:
The final expression of pH obtained after logarithms applied to both side. Buffer capacity is a measure of the ability to resist pH change and depends on both the absolute and relative component concentration. The buffer capacity is affected by the relative concentration of the buffer component which the buffer capacity in increasing with the concentration of the components of a buffer. For a given addition of acid or base, the buffer component concentration ratio change less when the concentration are similar than what they are different. Materials:
Acetic acid (MW=60 g/mol), NaOH solution (0.5M), HCl solution (0.001M), calibration buffer (pH3.5 and pH 5.5), 7-UP, 100 Plus. Apparatus:
pH meter, pipettes (10mL), volumetric flask (250mL), beakers (150mL), burettes, burette holder and stand, funnel, graduated cylinder (100mL), pipette bulb, weighing paper, spatula, stirring hot plate and stir bars.
1. The pKa of acetate buffer was given as 4.75, the amount of acetic acid and NaOH required preparing 250mL of 0.05M acetate buffer of pH 4.5 was calculated. 2. The above calculation was used to prepare 250mL of 0.05M acetate buffer of pH4.5. 3. The pH meter was standardized and the pH of the buffer was measured. 4. 200mL of buffer solution was titrated with 0.5M NaOH. The volume of NaOH which is needed to increase the pH of the buffer by 1 unit was recorded. 5. Step 4 was repeated by using 0.001M HCl, 7-Up and 100-Plus, in place of acetate buffer. Results
Volume of NaOH
Preparation of 250mL 0.05M acetate buffer
CH3COOH ↔ CH3COO- + H+
pKa Acetate buffer is 4.75, pH required is 4.5
[HA] is concentration of weak acid which is CH3COOH.
[A-] is concentration of conjugate base.
Concentration of CH3COOH:
Density of CH3COOH= 1.05kg/L
MW of CH3COOH= 60g/mol
Concentration of CH3COOH= Density/MW
4.5 = 4.75 – log ([CH3COOH]/[CH3COO-])
Concentration of acetate buffer required is 0.05M.
0.25 = log([CH3COOH]/[0.05])
[CH3COOH] = 0.0889M
To prepare 0.0889M [CH3COOH] from 17.5M CH3COOH,
Dilution factor is used, M1V1 = M2V2
(17.5)(V1) = (0.0889)(250)
V1 = 1.27mL
To prepare 0.05M conjugate base, NaOH is used to prepare conjugate base. NaOH + CH3COOH CH3COONa + H2O
As the reaction proceeds to complete, so concentration of conjugate base is equal to concentration of NaOH. Dilution factor is used to prepare 0.05M NaOH from 0.5M NaOH, M1V1 = M2V2 (0.5)(V1) = (0.05)(250)
V1 = 25mL
Calculation of buffer capacity,
0.05M Acetate buffer
Volume of NaOH used = 4.225cm3
Number of moles of NaOH used= 0.05 x 4.225/1000
= 2.1125 x 10-4 moles
100mL 0.05M Acetate buffer requires 2.1125 x 10-4 moles NaOH, 1L 0.05M Acetate buffer requires 2.1125 x 10-3 moles.
Buffer capacity= 2.1125 x 10-3M/1.015pH
= 2.081 x 10-3M/pH
Volume of NaOH used = 0.1cm3
Number of moles of NaOH used= 0.05 x 0.1/1000
= 5 x 10-6 moles
100mL 0.05M Acetate buffer requires 5 x 10-6 moles NaOH, 1L 0.001M HCl requires 5 x 10-5 moles.
Buffer capacity= 5 x 10-5 M/0.62pH
= 8.065 x 10-5M/pH
Volume of NaOH used = 1.875cm3
Number of moles of NaOH used= 0.05 x 1.875/1000
= 9.375 x 10-5 moles
100mL 0.05M Acetate buffer requires 9.375 x 10-5 moles NaOH, 1L 0.05M Acetate buffer requires 9.375 x 10-4 moles.
Buffer capacity= 9.375 x 10-4M/1.005pH
= 9.328 x 10-4M/pH
Volume of NaOH used = 2.3cm3
Number of moles of NaOH used= 0.05 x 2.3/1000
= 1.15 x 10-4 moles
100mL 0.05M Acetate buffer requires 1.15 x 10-4 moles NaOH, 1L 0.05M Acetate buffer requires 1.15 x 10-3 moles.
Buffer capacity= 1.15 x 10-3M/1.01pH
= 1.139 x 10-3M/pH
Calculation of theoretical buffer capacity of 0.05M acetate buffer and 0.001M HCl
(0.0889 - x)
(0.05 + x)
From Henderson – Hasselbalch equation,
pH= pKa - log( [CH3COOH]/[CH3COO‑] )
5.965= 4.75 - log( [CH3COOH]/[CH3COO-] )
5.965= 4.75 - log([0.0889 - x]/[0.05 + x])
-1.215= log([0.0889 - x]/[0.05 + x])
10-1.215= ([0.0889 - x]/[0.05 + x])
0.0610= ([0.0889 - x]/[0.05 + x])
0.0030 + 0.061x = 0.0889 - x
x = 0.0810M
As concentration of NaOH is equal to concentration of acetate buffer, No. of moles of OH- used in titration with 100mL 0.05M acetate buffer, N= 0.0810M (100/1000)L
No. of moles of OH- in 1L of 0.05M acetate buffer will be 0.081M. Theoretical buffer capacity of 0.05M acetate buffer = 0.081M/pH
% error = (Different in theoretical and experimental value / theoretical value) x 100%
= (0.081 - 2.081 x 10-3)/(0.081) x 100%
At pH 3.155, pH = -log [H+]
[H+] = 6.998 x10-4M
In 100mL, the number of moles = 6.998 x 10-5moles
When pH increases to 4.155, pH = -log [H+]
[H+] = 6.998 x10-5M
In 100mL, the number of moles = 6.998 x10-6M
Change in number of moles of HCL in 100mL= 6.998 x10-5moles - 6.998 x10-6moles
= 6.299 x 10-5moles
In 1L of 0.001M HCl, the changes in number of moles of HCl = 6.229 x 10-4moles. HCl + NaOH H2O + NaCl
Number of moles of NaOH requires is 6.229 x 10-4moles.
Theoretical buffer capacity = 6.229 x 10-4M/pH
% error = (Different in theoretical and experimental value / theoretical value) x 100%
= (6.229 x 10-4 - 8.065 x 10-5)/(6.229 x 10-4) x 100%
In this experiment, this acetate buffere solution is made up from weak acid (acetic acid) and strong base(NaOH). Acetic acts as week acid, it dissociates only slightly in water when dilution to 0.05M at pH 4.5. CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO-(aq)
When sodium acetate solution added, acetate ion and H3O+ ion from the acid enter the system. Based on Le Chatelier principle, the equilibrium of the system will shifts to the left to supress the acid from dissociating as much as it would in water and causes the [H3O+] decreased. Therefore higher pH (less acidic) obtined with increasing of the sodium acetate added. This is because the common ion effect that acetate ion CH3COO- play the role as common ion for both the acetate acid and the sodium acetate solution.
Based on the result obtained, the volume of NaOH needed to increase the pH of the buffer by 1 unit for acetate solution is the highest and for the 0.01M HCl is the lowest. Same case goes with their buffer capacity that have been calculated. This implies that the more concentrated the buffer, the greater its capacity, and the larger the resist of pH change. For the 7-Up and 100-Plus, citrate buffer is present in the solution which in the form of sodium citrate.
The percentage error obtained in this experiment is considerable high for both buffer and HCl acid solution. There are several factors that may be possible lead to the occurrence of error: 1. The solution is not homogeneous because not stirred well before adding sodium acetate 2. Systematic error maybe happened especially acetic acid measurement taking for 2 decimal places by using a 2mL pipette. For the titration between HCl and NaOH solution is strong acid and strong base titration. They will both fully dissociate, which mean all the molecules of acid and base will completely separate into ions. Therefore, the NaOH solution needed to change the pH of HCl solution is lesser than the titration of acetate buffer with NaOH solution. HCl + NaOH NaCl + H2O
The error of the experiment can be reduced by using micropipette for the 2 decimal place volume of acetic acid that need to take, instead of pipette which has 2mL. Besides, the glass rod may hit the pH meter bulb and it is unable to let the solution keep stirring, therefore the magnetic stirrer bar is recommended here to be used in order to have constant stirring to ensure the solution is homogeneous. Question and Answer:
1. Based on the results in the experiment, which beverage (7-Up or 100-Plus) shows a better buffer capacity? Why? Answer: In this experiment, 100-Plus has better buffering capacity than the 7-Up. Buffer capacity is the measure of this buffer ability to resist pH change and depends on both the absolute and relative component concentrations. The greater the buffer capacity of the buffer system, the more acid or base is required to change the pH of the buffer system. In this experiment, the average volume of NaOH solution used to increase 1 unit of pH is higher in 100-plus.
2. What are the chemical components in 7-Up and 100-Plus which are involved in determining the buffer capacity of these soft drinks? Answer: The chemical components involved in determining the buffer capacity of 7-up and 100-plus are citric acid, its conjugate base, citrate, carbon dioxide and its conjugate base carbonic acid. CH3COOH ↔ CH3COO- + H+
CO2 + H2O ↔ H2CO3
Buffer work through a phenomena known as common-ion effect. The common-ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of equilibrium mixture shift away from forming more of it. Therefore buffer systems able to resist the pH change when a small amount of acid or base is added and on dilution. Buffer capacity is the measure of this buffer ability to resist pH change and depends on both the absolute and relative component concentrations. The buffer capacity of 0.05M acetate buffer is 2.081 x 10-3M/pH, 0.001M HCl is 8.065 x 10-5M/pH, 7-Up is 9.328 x 10-4M/pH and 100-plus is 1.139 x 10-3M/pH. The error for both acetate and HCl are higher than theoretical value which are 97.43% and 87.20% respectively.
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