Acid Dissociation Constant of Methyl Red
Abstract: This report deals with the experiment Acid Dissociation Constant Of Methyl Red. Using the Handerson-Hasselbach equation as well as the values obtained with the aid of a spectrophotometer, pKa and Ka values were determined. In this experiment, three important steps were performed to come up with the necessary values. The first step involved the determining of the wavelength at which HMR and MR exhibit absorption maxima using two solutions.The second step involved determining the absorbance indices at wavelength maximum for the acid and base. The Beer’s law was also verified for both the HMR and MR. The last step involved the determination of the HMR and MR amounts present in solution as a function of pH.
Introduction: The Acid Dissociation Constant of Methyl Red experiment aims to obtain the pKa and Ka values for methyl red. These values will tell us the concentrations were in methyl red will be at equilibrium. Methyl red will be very vital in this experiment because it will be the main chemical used together with a spectrophotometer. The said chemical would be used because it gives high absorption peaks in the visible portion of the specrtrum. The main equation to be used to obtain the desired values is the equilibrium constant
equation
and
the
Handerson-Hasselbach
equation.
They
are
given
by
Ka=([H][MR])/([HMR]) and pKa=pH-log[MR]/[HMR] respectively.HMR and MR valus that were obtained in brackets would indicate that these values must be in molarity(mol/L). HMR and MR concentrations can easily be obtained from the values that the spectrophotometer would produce after conducting the experiment. Using Beer’s Law, A=abc, a=molar absorbancy, b=length of lightpath and c= concentration and the relationship A=a1HMR[HMR]+a1MR[MR] and A=a2HMR[HMR]+a2MR[MR], simple manipulation of the equations would lead to the values of HMR and MR.
Materials and methods: To perform the experiment, the following will be needed: pipet, volumetric flask, cuvette, spectrophotometer and a pH meter. As stated earlier, three necessary steps will have to be done. The first step includes preparation of solutions A and B. Solution A which was prepared by dilution of 10mL of standard methyl red and 10mL of 0.1 HCl to 100mL and solution B which was prepared by dilution of 10mL of methyl red and 25mL of 0.4 M NaOAC to 100mL, were placed at matched cuvettes to measure the absorbance versus wavelength between 350 and 600 millimicrons. The second step-portions of A and B were diluted to 0.75, 0.5 and 0.25 times their initial concentrations using 0.01 M HCl and 0.01 M NaOAC respectively. Water, as the blank solution versus the absorbance of these solutions was made. Third step- Series of solutions were made by adding amounts of 0.02 acetic acid to constant amounts of standard indicator buffered with 0.04 M acetic acid. After which, absorbances of the solution and pH values were determined. Results and discussions: Figure 1
Figure 1 is a graph obtained by using the spectrophotometer. This graph shows the wavelengths where solutions A and B have maximum absorbances. For solution A, the graph reads that the maximum absorbance is at 519.20 nm. For solution B, the graph reading states that the maximum absorbance is at 398.20 nm. This two values will be essential for determining the concentrations of HMR and MR in the latter part. Figure 2.
0.16 0.14
absorbance
0.12 0.1
sol A at Amax
0.08
sol A at Bmax
0.06
sol B at Amax
0.04
sol B at Bmax
0.02 0 -0.02
0
0.5
1
1.5
composition
The graph shown in figure 2 is the graph composed of several lines. These lines are readings of a particular solution at a particular wavelength and particular solution. Legends are shown along with the graph and solutions used are the dilutions made to 0.75, 0.50, and 0.25. It can be easily seen that absorbance is directly proportional with concentration. The values of a1HMR, a1MR, a2HMR and a2MR can also be obtained from these graphs through their slopes. With all the necessary values at hand, we can now calculate for the concentrations of HMR and MR and eventually calculate for Ka and pKa. Sample Calculation Given: A = 519.20 a at A: y = 0.15020x – 0.00842
B = 398.20
a at B: y = 0.01245x + 0.00610 b at A: y = 0.1396x – 0.00226 b at B: y = 0.06450x + 0.00238 using equation A1 = a1HMR[HMR] + a1MR-[MR-] A2 = a2HMR[HMR] + a2MR-[MR-]
such that a = m(slope) 143
By substitution: 519.20 = 0.15020[HMR] + 0.01245[MR -] 398.20 = 0.01396[HMR] + 0.0.06450[MR-] [HMR]= 2298.792949 [MR-]= 5524.602332 With the pH values at hand, which are 5.10 for 10mL, 4.85 for 20mL and 4.63 for 30mL, we can calculate for pKa. pKa = pH – log [MR-]/[HMR] pKa = 5.10 – log (5524.60/2298.79) pKa = 6.92 Ka = 1.91 x 10-5 Conclusion: After performing the experiment and making the necessary calculations, it can be concluded that HMR and MR have values equal to 2298.8 and 5524.6 respectively. The average Ka and pKa constants have values of 5.21 and 3.64x10-5 respectively. Reference: Alberty, R.A., Silbey, B.J. Physical Chemistry; John Wiley and Sons, Inc: New York, 1995, p.271. http://en.wikipedia.org/wiki/methyl_red (accesed Jan. 22, 2011)
