exp5 520

Abstract: The objective of this experiment is to determine the degree of dissociation, α and the thermodynamic equilibrium constant, Kα for acetic acid, CH3COOH at 25oC. To prepared the acetic acid solutions, the pipettes and the volumetric flasks are required. Firstly, fill deionized water into the volumetric flasks containing 0.1M of acetic acid up to the calibration mark to prepare solutions of CH3COOH with molarities 0.05, 0.01, 0.005, 0.001, 0.0005 and 0.0001M. Now we have six acetic acid solutions available for measuring the conductivity at different concentrations. Then, calibrated the digital conductivity meter with conductivity probe throughly with deionized water. After thoroughly rinsing the conductivity probe with deionized water, fixed it in the clamp, so that its height can be adjusted by loosening and fastening the screw of the crossed bosshead at the stand. Started with the solution of the lowest concentration 0.0001M and following the order up to concentrated acetic acid, the value of

was measured. The experimental value of dissociation constant of acetic acid Ka is

1.8197x10-3 and the experimental value of molar conductivity of acetic acid ʌ˳ is 125 Scm2mol-1. Meanwhile the degree of dissociations, α, at concentrations 0.05M, 0.01M, 0.005M, 0.001M, 0.0005M and 0.0001M is 0.077, 0.176, 0.240, 0.720, 0.480 and 0.800. We succeed to obtain the objective of this experiment.

Introduction: Kohlrausch’s law describing the effect of dilution on the molar conductivity of strong electrolytes, Kohlrausch’s law of independent ion migration and the Debye‐Hückel limiting law (both in general and simplified forms) and explain their importance. The conductivity meter actually measures a solution’s resistance but displays its conductivity. Explain how a set of conductivity standard solutions can be used to calibrate the conductivity meter and how the meter can then calculate conductivity, 𝜅,values (your explanation should include the concept of a cell constant). Then define equivalent conductance, Λ, as:

Eq1 Here, v is the number of equivalent ions per molecule, α is the fractional dissociation, U is the ionic mobility of the charged species and c is the concentration of the electrolyte in mol/L. Using this formula will yield ʌ having unit Ω-1m2mol-1 (or S-1 m2mol-1). For strong electrolytes

is unity and ʌ is roughly constant at all concentrations. It will however

approach a finite value at infinite dilution, ʌ. Onsager showed that for strong electrolytes in dilute solution as:

˳(

√ )

Eq2

Thus, the conduvtance at infinite dilution of solutions of strong electrolytes may be detremined from measurements of conductance at varying concentration. The y-intercept of a graph of ʌ versus √ will yield this value. For weak electrolytes, the fraction dissociated α is not unity but is given by the ration of equivalent conductance to the conductance at the infinite dilution, assuming that ion mobilities are independent of concentration

Eq3

Now present the expression for the equilibrium constant, 𝐾c with c expressed as a molarity: 𝐾

Eq4

The equilibrium constant for the dissociation of acetic acid obtained using this equation will differ from 𝐾α, the true thermodynamic value, because the activity coeffeicient, γ, are left out of the numerator and because of the assumptions made when calculating α using equation 3. 2 Since Kα = 𝐾cγ ± is a good approximation, it follows that:

log 𝐾𝑐

log 𝐾𝛼

𝑙𝑜𝑔𝛾±

𝐸𝑞5

According to the Debye-Huckel Theory, the mean activity coefficient at low and moderate ionic concentration is given by: √

±

Eq6

√

The ionic strength is defined by the expression:

∑

Eq7

Where 𝑚 is the molality of the th ionic species in moles per kilogram of solvent, is the charge on ion and the sum is taken over all ionic species in the solution. In equation 6, for dilute solutions at 25oC, A= 0.509kg1/2mol-1/2 and the quantity B is nearly unity for many electrolytes. In addition, it is valid to approximate molality by molarity(mol/L). For the very dilute solutions of acetic acid used in the experiment, the I=αc is very small and equation 6 is well approximated by:

log 𝐾

log 𝐾

5

√

Hence, if Kc has been determined at many dilute concentrations, a plot of log Kc versus √ will give a straight line. Extrapolation to c=0 (the y-intercept) can be made to yield log Ka from which Ka can be easily calculated.

Procedure: 1. 2. 3. 4. 5. 6. 7. 8.

Cleaned burette with 0.1M CH3COOH solution. Drained out required volume of 0.1M CH3COOH into six volumetric flasks. Added deionized water to the calibration mark for each volumetric flask to prepared molarities of 0.05M, 0.01M, 0.005M, 0.001M, 0.0005M and 0.0001M. Calibrated the digital conductivity meter with conductivity standard of 84μS cm-1 at 25oC. Rinsed stirring bar with 50mL deionized water and put it in the 50mL beaker with deionized water and placed the magnetic stirrer. Immersed probe 5cm in the solution and switch on the magnetic stirrer. Record the electric conductivity к Repeated step 5 to 7 with diluted CH3COOH. Begin with the lowest molarity to the higher molarity of CH3COOH.

Results: Data sheet experiment 5: Concentration (M) 0.05 0.01 0.005 0.001 0.0005 0.0001

Electrolytic conductivity of HAc (к) at 25oC (μS cm-1) 480 220 150 90 30 10

Electrolytic conductivity of deionized water (кDIwater) at 25oC = 1.9μS cm-1

Calculation and discussion: The value of ʌ(S cm2 mol-1) for CH3COOH solutions is: i)

C = 0.0001M К = 10 μS cm-1

𝑚

𝑚

𝑚 𝑚

𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

ii) C = 0.0005M К = 30 μS cm-1

𝑚

𝑚

5 5

5

𝑚 𝑚

𝑚 𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

5

𝑚 𝑚

iii) C = 0.001M К = 90μS cm-1

𝑚

𝑚

𝑚 𝑚

𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

iv) C = 0.005M К = 150 μS cm-1

5

𝑚

5

𝑚

5 5

5 5

𝑚 𝑚

𝑚 𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

5

𝑚 𝑚

v) C = 0.01M К = 220μS cm-1

𝑚

𝑚

𝑚 𝑚

𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

vi) C = 0.05M К = 480 μS cm-1

𝑚

𝑚

5 5

5

𝑚 𝑚

𝑚 𝑚 𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝑚

5

𝑚 𝑚

ʌ (S cm2 mol-1)

1/ʌ (S-1 cm-2 mol)

C (mol cm-3)

100.0 60.0 90.0 30.0 22.0 9.6

0.010 0.016 0.011 0.033 0.045 0.104

0.01x10-5 0.05x10-5 0.10x10-5 0.50x10-5 1.00x10-5 5.00x10-5

cʌ

(S cm-1 mol) 1.00x10-5 3.00x10-5 9.00x10-5 15.0x10-5 22.0x10-5 48.0x10-5

From this experiment, we obtained the values of ʌ by using the formula molar conductivity, which is 100, 60, 90, 30, 22 and 9.6 Scm2mol-1. Through this, we also can calculate the values of 1/ʌ which is 0.010, 0.016, 0.011, 0.033, 0.045 and 0.104 S-1cm-2mol. Meanwhile the values of cʌ that we obtained are 1.00x10-5, 3.00x10-5, 9.00x10-5, 15.0x10-5, 22.0x10-5, 48.0x10-5. From this resulted, we can plotted graph 1/ʌ versus cʌ, the graph used to determine the value of ʌ˳ by extrapolation to zero concentration. The equation c = 1/ ʌ˳ implies that, if 1/ ʌ˳ is plotted against cʌ, then the intercept at c=0 will be 1/ ʌ˳. Value of 1/ʌ˳ = 0.008 S-1cm-2mol

ʌ˳= ʌ = 125 Scm2mol-1 Then, we can calculate the value of Kc from the slope of the graph by using formula:

This equation represents Y = C + MX: Y = 1/ʌ

C = 1/ʌ˳

M=

From the graph slope, we find that value of M written as:

5 5

X = cʌ

Substitute in: M= 585.71x105 = ]

𝐾

585.71 𝐾

]

𝐾

]

5

5 5

√

𝐾

𝐾 𝐾 𝐾

S-1cm-1

According to the standard data of molar ionic conductivity, the theoretical value of molar conductivity in the limit of zero concentration of acetic acid:

˳ 5 5

𝑚 𝑚 𝑚 𝑚

From calculation, the experimental value of ˳ is 125 Scm2mol-1 (from Kohlrausch’s law). |

5

5 𝑚 𝑚 5 𝑚 𝑚

|

The major systematics error in this experiment is not rinsed the electrode with deionized water properly before use. This cause the contaminant in the smaple to be tested and lead to higher percentage error which is 67.99% and have low accuracy between experimental and theoretical results. On the other hand, calculation of degree of dissociation (α) of CH3COOH at the concentrations of 0.05M, 0.01M, 0.005M, 0.001M, 0.0005M and 0.0001M is: At 0.05M(5.00x10-5molcm-3)

Λ = 9.6

α = Λ / Λ˳ = 9.6 / 125 = 0.077

At 0.01M(1.00x10-5molcm-3)

Λ = 22.0

α = Λ / Λ˳ = 22.0 / 125 = 0.176

At 0.005M(0.50x10-5molcm-3)

Λ = 22.0

α = Λ / Λ˳ = 30.0 / 125 = 0.240

At 0.001M(0.10x10-5molcm-3)

Λ = 22.0

α = Λ / Λ˳ = 90.0 / 125 = 0.720

At 0.0005M(0.05x10-5molcm-3)

Λ = 22.0

α = Λ / Λ˳ = 60.0 / 125 = 0.480

At 0.0001M(0.01x10-5molcm-3)

Λ = 22.0

α = Λ / Λ˳ = 100.0 / 125 = 0.800

Degree of dissociation, α is inversely proportional to concentration of given electrolyte, thus as the concentration decreases, the degree of dissociation increases. As the concentration C approaches to zero or dilution approaches to infinity, the degree of dissociation, α reaches to unity that is maximum value of that. This generalization is called as Ostwald dilution law. According to Physical Chemistry (8th edition), thetheoretical value of dissociation constant of acetic acid (Ka) is 1.4 x 10-5M. From calculation, the experimental value of Ka is: log 𝐾

log 𝐾

log

5

log 𝐾 5

√ 5

5

√

log 𝐾 log 𝐾

𝐾

𝑚

𝑚

𝐾

|

𝑚 𝑚

𝑚

𝑚

|

The molar conductivity is found to vary according to the concentration. One reason for this variation is that the number of ions in the solution might not be proportional to the concentration of the electrolyte. For instance, the concentration of ions in a solutionof a weak acid depends on the concentration of the acid in acomplicated way, and doubling the concentration of the acid added does not double the number of ions. Secondly, because ions interact strongly with one another, the conductivity of a solution is not exactly proportional to the number of ions present.

Conclusion: In conclusion, we conclude that the experimental value of dissociation constant of acetic acid Ka is 1.8197x10-3 and the experimental value of molar conductivity of acetic acid ʌ˳ is 125 Scm2mol-1. Meanwhile the degree of dissociations, α, at concentrations 0.05M, 0.01M, 0.005M, 0.001M, 0.0005M and 0.0001M is 0.077, 0.176, 0.240, 0.720, 0.480 and 0.800. We succeed to obtain the objective of this experiment.