Transcript

EXPERIMET 2 : DETERMINATION OF THE VALUE OF A WEAK ACID

SUMMARYThis experiment is conducted to identify a weak acid by determining its Ka value by titration. p value is determined from titration curves. The acid strength of a weak acid is measured by the dissociation constant, . The larger the value of , the stronger the acid is. In this experiment, 40mL of an unknown acid is titrated with 0.1M of sodium hydroxide solution. Two times titration is carried out and the results obtain is expressed in titration curves by plotting a graph of pH value against volume of sodium hydroxide (NaOH) solution. INTRODUCTION

Acid and bases are usually being classified into strong or weak. However, for acid values are commonly used. In this experiment, the aim is to identify a weak acid by determining its value via titration. Titration is done by a standard sodium hydroxide solution. The obtained titration plot is used to determine the dissociation constant () unknown. The ionization of a weak acid can be expressed by the following equation:HA(aq)H+(aq)+A-(aq)(Eq. 1)Since an equilibrium exists, an equilibrium constant, Ka, can be written:Ka=[H+] [A-](Eq. 2) [HA]

The value is an indication of the acids strength. The larger the , the stronger the acid is. This characteristics is use to identify the unknown acid. A similar system exist for bases,

Titration method is use in this experiment. A sample of an unknown acid is titrated with 0.1M of sodium hydroxide solution. The pH values are recorded every mL of NaOH titrated. The pH values are plotted against the volume of NaOH solution used. From the graph, the equivalence is determined. Then the volume of base halfway to the equivalence point is obtained and the pH value at this volume is recorded. The [H+] corresponding to this pH is equal to the Ka for the acid. At a point halfway to the equivalence point, [H+] = [HA] = [A-] for a monoprotic acid. Canceling out [A-] and [HA] in Equation 2 gives Ka = [H+]. AIM

The objective of this experiment is to observe and measure a weak acid neutralization and determine the identify of an unknown acid by titration.

THEORY

The purpose of this experiment is to identify an unknown weak(monoprotic) acid by titration with a standard sodium hydroxide solution. The pH of the titration solution will be monitored using a pH meter, the obtained titration plot will then be used to determine the equivalent molecular weight and dissociation constant (Ka) of the unknown. These values will be used to determine the identity of the weak acid. When a weak acid (HA) is dissolved in water, only some of the molecules will dissociate to yield H3O+ and A- ions. At this point, a dynamic equilibrium is established:

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Under these equilibrium conditions, the total concentration of each species remains constant, even though the species in solution are constantly dissociating and recombining. The degree of dissociation of the weak acid is used to characterize the acid, and is calculated according to the equation:

pKa = -log [Ka],

In the expression, Ka is the acid dissociation constant. Strong acids typically dissociate completely, and therefore would have a Ka value of greater than 1. Weak acids have Ka values much smaller than 1 (typically less than 10-4). For example, the Ka of acetic acid (vinegar) is 1.75 10-5, while the Ka of bicarbonate (baking soda) is 4.7 10-11. For convenience, scientists often use the pKa of weak acids, as it allows them to work with whole numbers (pKa = -log Ka). The pKa values of acetic acid and bicarbonate are 4.75 and 10.32, respectively.

When a strong base is added to a solution of a weak acid, the hydroxide ion reacts with some of the H3O+ present, therefore disturbing the equilibrium. More of the acid will dissociate, until a new equilibrium is established. When the number of moles of base added equals the number of moles of weak acid present, a sharp change is observed in pH, which can be detected using either a visual indicator or pH meter. This point is the equivalence point, which is is theamount the base needed to neutralize all of the original monoprotic acid and any additional base added simply increases the pH.

The sudden change in the solution shows that titration has reached the equivalence point. pH in aqueous solution is related to its hydrogen ion concentration. Symbolically, the hydrogen ion is written as [H3O+]. pH is defined as the negative of the logarithm of the hydrogen in concentration. This information can be used to determine the quantity (in moles) of acid that is present. pH = -log [H3O+]

The volume and concentration of the added base can be used to determine the number of moles of acid present (by assuming a 1:1 molar ratio of acid : base). By measuring the pH of solution after each addition of base, a titration curve can be constructed. The titration curve allows for the determination of the Ka value of the acid. According to the equation above, Ka will be equal to [H3O+] when [A-] = [HA], and the pKa will equal the pH at this point. This condition is satisfied halfway to the equivalence point of the titration. A graph of pH versus mL of NaOH added can be drawn by carefully following the titration with a pH meter. A pH meter is a highly sensitive instrument than converts voltage caused by H+ in solution to pH readings. A titration curve has a very distinct and offers considerable information about the acid/base in question. The molar mass of the monoprotic acid can be determined from the equivalence point. Since one mole of NaOH will neutralize one mole of monoprotic acid, at the equivalence point, the following relationship holds:

Vbase x Mbase = moles of base = moles acidMMacid =

The pH at the equivalence point depends on the extent of hydrolysis of the conjugate acid/base produced in the neutralization reaction. For example, consider the reaction of the weak acid, HA, with NaOH: HA + OH- ------> H2O + A

At the equivalence point, all of the original HA is neutralized, leaving only [A-] in solution. This A- will than hydrolize to produce a slightly basic solution: A + H2O HA + OH

This titration curve may also be used to obtain pKa and Ka. Consider, again, the titration of a weak acid, HA, with a strong base. Because HA is a weak acid, there will be some HA present in equilibrium with A-, at every point in the titration. The equilibrium concentrations are, of course, related by the acid dissociation constant, K.HA H+ + A- Ka =

This equation may be converted to the logarithmic form, and rearranged, to give the Henderson Hasselbalch Eq:

pH = pKa + log

This equation shows that the pH of a weak acid depends upon the pKa of the acid, and upon the ratio of the concentrations of the anion and undissociated acid. When the acid is half neutralized, [HA] = [A-], so the logterm cancels out and pH at this point equals pKa. Since pKa = -log [Ka], we can solve for Ka.APPARATUS

Burette, pipette, conical flask, beakers, retort stand with clamp, wash bottles, volumetric flask, pH meter, beaker

MATERIALS

0.1 M unknown acid (monoprotic acid), 0.1 M sodium hydroxide solution, distilled water

PROCEDURE

1. 0.1M of sodium hydroxide solution, NaOH is transferred in a burette. 2. 40mL of unknown monoprotic acid solution is prepared in a volumetric flask. 50mL of distilled water is added to the unknown monoprotic acid so that the pH electrode will be covered.3. The initial reading of the NaOH solution in the burette is recorded.4. The pH electrode is rinsed and it is placed in the beaker containing monoprotic acid.5. 1mL of NaOH solution is added and the pH value of the solution is recorded after each addition.6. When the pH value has reached to 5.5, 0.5mL of NaOH solution is added and the NaOH is continue added until the pH is in constant.7. step 1 to 6 is repeated once more for the second titration.8. A graph of pH value against volume of NaOH solution is plotted. The equivalence point of each graph is determined. The pH value of base at this time is converted to [] to give thevalue of acid.

RESULTS AND CALCULATION.Titration 1I. NaOH initial = 50mLII. Unknown monoprotic acid = 40mL + 50mL distilled waterIII. Table :Volume of NaOH used (mL)pH valueVolume of NaOH used (mL)pH value

0.04.0130.55.62

1.04.0631.05.66

2.04.1431.55.71

3.04.2232.05.75

4.04.3032.55.79

5.04.3733.05.84

6.04.4333.55.90

7.04.4934.05.95

8.04.5434.56.01

9.04.6035.06.08

10.04.6535.56.16

11.04.7036.06.26

12.04.7536.56.39

13.04.7937.06.54

14.04.8437.56.79

15.04.8838.07.32

16.04.9238.59.57

17.04.9739.010.32

18.05.0139.510.57

19.05.0540.010.75

20.05.1340.510.88

21.05.1541.010.98

22.05.1842.011.13

23.05.2143.011.24

24.05.2644.011.32

25.05.3145.011.39

26.05.3646.011.45

27.05.4247.011.50

28.05.4748.011.54

29.05.5449.011.58

29.55.5550.011.62

30.05.58

Graph of pH value versus Volume of NaOH used (mL)

Equivalence point = 38.50 mL , Half-equivalence point = 19.25 mL

Sample calculation titration 1

I. Molarity of unknown acid :

Moles of NaOH solution = 0.1mol x 19.25mL x = 1.925x10-3 moles

From the reaction equation, number of moles of base is equivalent to number of moles of acid. Therfore, n,Base = n,Acid

Moles of unknown = 1.925x10-3 moles x = 1.925x10-3 moles solution, HA

Prepared solution of HA and NaOH solution is 0.1 M

II. The acid ionization constant Ka :

pH at half-equivalent point = pKA = 5.11

pKa=-log Ka5.11=-log KaKa=105.11=7.76 x 106

Titration 2I. NaOH initial = 50mLII. Unknown monoprotic acid = 40mL + 50mL distilled waterIII. Table :Volume of NaOH used (mL)pH valueVolume of NaOH used (mL)pH value

0.04.0430.55.58

1.04.1031.05.62

2.04.1731.55.65

3.04.2432.05.69

4.04.3132.55.72

5.04.3733.05.77

6.04.4333.55.81

7.04.4934.05.91

8.04.5434.55.97

9.04.5935.06.04

10.04.6435.56.11

11.04.6936.06.19

12.04.7436.56.29

13.04.7837.06.41

14.04.8237.56.58

15.04.8738.06.79

16.04.9138.57.29

17.04.9539.09.16

18.05.0039.59.77

19.05.0540.010.16

20.05.1340.510.46

21.05.1641.010.66

22.05.1941.510.81

23.05.2342.010.93

24.05.2642.511.00

25.05.3143.011.08

26.05.3544.011.19

27.05.4145.011.29

28.05.4646.011.35

29.05.5247.011.41

29.55.5548.011.46

30.05.5849.011.51

50.011.55

Graph of pH value versus Volume of NaOH used (mL)

Equivalence point = 38.76 mL , Half-equivalence point = 19.38 mL

Sample calculation titration 2

I. Molarity of unknown acid :

Moles of NaOH solution = 0.1mol x 19.38mL x = 1.938x10-3 moles

From the reaction equation, number of moles of base is equivalent to number of moles of acid. Therfore, n,Base = n,Acid

Moles of unknown = 1.938x10-3 moles x = 1.938x10-3 moles solution, HA

Prepared solution of HA and NaOH solution is 0.1 M

II. The acid ionization constant Ka :

pH at half-equivalent point = pKA = 5.10

pKa=-log Ka5.10=-log KaKa=105.10=7.94 x 106

Average Ka from titration 1 and titration 2 :

Average Ka =

= =7.85 x 106

Sample error calculation.

From the calculation, the unknown monoprotic acid is assumed to be either acetic acid or potassium hydrogen phthalate. The theoretical value of pKa value of acetic acid is 4.756 and for potassium hydrogen phthalate is 5.406. Ka value of acetic acid is 1.75 x 105 and for KHP is 3.93 x 106 . However, the results obtained from titration, the value of pKa obtained is 5.11 whereas the Ka value is 7.85 x 106. By calculating the percentage error, the acid can be determine.

For acetic acid :Percentage error pKa = x 100% = 7.44 %

Percentage error Ka = x 100% = 55.14 %

For KHP :

Percentage error pKa = x 100% = 5.51 %

Percentage error Ka = x 100% = 99.83 %

Discussion

The main objective of this experiment is to identify a weak acid by determining its Ka value by titration. Based on the experiment, graphs are build to represent the titration data obtained. The experiment are conducted twice, titration 1 and titration 2 to get the most accurate data by calculate the average. The equivalence points for titration 1 is 38.50 mL while for titration 2 is 38.76 mL. The half way volume of titration which is half equivalence point for titration 1 is 19.25 mL and the pH of the solution is 5.11 while, for titration 2 is 19.76 mL and the pH is 5.10. From the calculations, the average Ka value obtained is .

From the graph obtained for both titrations, the titration curve shows the characteristic features of weak acid-strong base titration. The initial curve shows the indicating a relatively low hydronium ion concentration created by the incomplete dissociation of a relatively weak acid. The flat portion of the titration curve before the equivalence point is called the buffer region. The pH between this point rises steadily during titration which reflecting the buffering activities. In the middle of the buffer region lies the half-equivalence point. Here, the volume of NaOH added is half the required to reach the equivalence point and half of the unknown monoprotic acid has been converted to conjugate base. So, at half of equivalence point, the pH will be equal to the pKa of unknown weak acid.

However, there is error that occur during conducting experiment. One possible error for the differences between both titration is the incorrect volume of NaOH used. The volume used is high which resulting in incorrect high calculated molarity. For example, the reading of NaOH in the burette should be 0.00mL, but in the experiment, the volume used does not same as the preferred one. The existence of air bubbles at the tip of burette also affect the volume of NaOH used in titration. This will affecting the result in calculation of molarity.

The another possible error is the solution of acid or base used might be defective. This occurred due to the contaminated apparatus used. The solution may contained impurities and lead to the presence of interfering substances in the standard solution of titration. As the result, the pH of the titration also interrupted by the impurities.

As the using of pH meter is more accurate than using phenolphthalein solution as indicator, the errors must be considered too. During the titration, the solution is titrated directly to the pH electrode. This will affect the pH of the solution. The solution should be titrated directly to the unknown acid solution in volumetric flask. Besides, the pH electrode does not perfectly rinsed with distilled water which resulting the incorrect pH reading.

CONCLUSION

From the experiment, the unknown acid is identified by titration process. The unknown acid is titrated with sodium hydroxide solution, NaOH. The graph is plotted to determine the equivalence point and half-equivalence point thus the pKa value can be determine which is equal to the pH value. From the calculation, the half-equivalence point determined was 19.31 mL and the pKa value is averaging at 5.11 .Theoretically value of pKa for acetic acid is 4.756 and potassium hydrogen phthalate is 5.406. Based from the Ka value, the value of acetic acid is 1.75 x 105 whereas the value for potassium hydrogen phthalate is 3.93 x 106 . From the calculations, the value Ka of unknown acid is 7.85 x 106 and shows in the rage of potassium hydrogen phthalate. By comparing between acetic acid and potassium hydrogen phthalate, it can be concluded that the unknown acid is likely to be potassium hydrogen phthalate due to the closer value of pKa which is 5.406. Thus it is known that the unknown acid is potassium hydrogen phthalate.

RECOMMENDATION

From the experiment, there are points of recommendation and precautions should be taken in order to get more accurate and efficient results.Firstly the standard solution should that is should be a pure without any other substances or contamination. The presence of other substances will affect the results. It should also be stable at room temperatures. Therefore, dried standard material before weighing and diluting is preferable.Indicator can also be used during the experiment. This will get a more faster results as the solution will change in color after the reaction has occur. Thus, the titration can be stopped immediately. This will save more time and avoid further reaction between the solution.Besides, the molecular weight of the unknown acid should also be considered. This is to compared the value between the unknown acid and other acid sample. Therefore, the mass of solid used for making the solution should be known as well.

REFERENCES1. http://web.utk.edu/~kcook/319S02/exp5m.pdf2. https://www.apsu.edu/sites/apsu.edu/files/chemistry/F12__Determining_the_Identity_of_an_Unknown_Weak_Acid.pdf3. Adapted from R. C. Kerber et. al http://www.sinc.sunysb.edu/Class/orgolab/che199_susb014.PDF; W.F. Kinard et.al 4. http://www.cofc.edu/~kinard/221LCHEM/2002CHEM221LabSchedule.htm; and D.C. Harris Quantitative Chemical Analysis, 5th ed Freeman 5. Press, 1996. http://www.dartmouth.edu/~chemlab/chem3-5/acid1/full_text/chemistry.html7. https://mymission.lamission.edu/userdata%5Cpaziras%5CChem102%5CExp_10.pdf8. Chang. R, 2010, Chemistry, 10 th Edition, United States : New York.Theodore L. Brown, H. Eugene LeMay, Jr., Bruce E. Bursten, Catherine J. Murphy, Patrick M. Woodward, 2012,Chemistry The Central Science, 12th Edition, United States9. http://www.heacademy.ac.uk/assets/ps/documents/database_of_praticals/determination_of_the_dissociation_constant_of_a_weak_acid_by_titration.pdf 10. Determination of The Dissociation Constant of a Weak Acid by Titration by Dr Barry O Grady, December 200311.

15