"experiment: spectrophotometric determination of the acid dissociation constant of methyl red"
1. Why is pH plotted against log([MR-]/[HMR]) to get the pKa of methyl red?
2. Why is it right that HMR has a pH equal to 2 and MR- to have a pH equal to 8?
1. pH is plotted against log([MR-]/[HMR]) to determine the pKa of methyl red because the pKa represents the pH at which half of the indicator molecule is in its acidic form (HMR) and the other half is in its basic form (MR-). By plotting pH against the logarithm of the ratio ([MR-]/[HMR]), we can observe the point at which this ratio is equal to 1, which corresponds to the pKa.
To get this ratio, you will need to perform an experiment. Here's a step-by-step guide on how to determine the pKa of methyl red using spectrophotometric methods:
1. Prepare a series of buffer solutions with different known pH values, spanning a range of acidic to basic pH values. Ensure that the buffer solution does not absorb at the wavelength of interest.
2. Measure the absorbance of each buffer solution at a specific wavelength that corresponds to the maximum absorbance of methyl red.
3. Prepare a solution of methyl red by dissolving it in an appropriate solvent.
4. Take a small volume of the methyl red solution and add it to each of the buffer solutions. This will create a series of solutions with known concentrations of the acidic form (HMR) and basic form (MR-) of methyl red.
5. Measure the absorbance of each solution at the same wavelength as before.
6. Determine the concentration of HMR and MR- in each solution by calculating their ratios using the Beer-Lambert law.
7. Plot the pH on the x-axis and the logarithm of the ratio ([MR-]/[HMR]) on the y-axis. The pKa of methyl red can be determined by identifying the pH value at which the ratio is equal to 1. This corresponds to the point where the curve intersects the value of log(1) on the y-axis.
2. The reason why HMR has a pH equal to 2 and MR- has a pH equal to 8 is due to the nature of the acidic and basic forms of methyl red in solution.
The acidic form, HMR, exists as a molecule that is protonated and positively charged, hence it is more likely to be present in an acidic environment. Therefore, at a pH value of 2 (an acidic condition), more of the indicator molecule will be in the acidic form.
On the other hand, the basic form, MR-, exists as a deprotonated molecule that carries a negative charge. This makes it more likely to be present in a basic environment. Therefore, at a pH value of 8 (a basic condition), more of the indicator molecule will be in the basic form.
The transition from the acidic form to the basic form occurs in the vicinity of the pKa value, which in the case of methyl red is around pH 5. At this pH, equal amounts of the acidic and basic forms are present, resulting in the ratio [MR-]/[HMR] being equal to 1.