1-Hydroxypyrene-3,6,8-trisulfonate (HPTS) is a monoprotic acid that can be used as a fluorescent aci — Acid-Base Chemistry / Analytical Chemistry Chemistry Question
Problem Context
1-Hydroxypyrene-3,6,8-trisulfonate (HPTS) is a monoprotic acid that can be used as a fluorescent acid-base indicator because its conjugate acid form does not emit light (on excitation at λ = 454 nm), while its conjugate base form does. The fluorescence intensity (at 520 nm) of an HPTS solution was measured in seven different phosphate buffers of differing pH, giving the relative fluorescence intensities shown below.
How much energy is lost as heat when a photon with λ = 454 nm is absorbed by HPTS and a photon with λ = 520 nm is emitted?
Model Answer
E = hc/λ, so
E = (6.626 × 10-34 J s)(2.998 × 108 m s-1)/(454 × 10-9 m) = 4.38 × 10-19 J excitation
E = (6.626 × 10-34 J s)(2.998 × 108 m s-1)/(520 × 10-9 m) = 3.82 × 10-19 J emission
Energy lost = 4.38 × 10-19 J – 3.82 × 10-19 J = 5.6 × 10-20 J
(5.6 × 10-20 J)(6.022 × 1023 mol-1) = 34 kJ mol-1
What is the pKa of HPTS? Explain your answer.
Model Answer
At pH = pKa, half of the HPTS will be in the conjugate base form and so the fluorescence will be at half-maximal intensity. Inspection of the curve gives the half-maximal fluorescence at pH = 7.4.
The phosphate buffers used in this experiment were prepared by adding solid NaOH to 100 mL of a 0.100 M solution of H3PO4. How many moles of NaOH would be required to make buffers with each of the following pH values? You may assume that the volume remains 100 mL. H3PO4 has pK1 = 2.12, pK2 = 7.21, and pK3 = 12.32.
pH = 6.50
Model Answer
Since 6.50 >> 2.12, almost all the first proton of H3PO4 will have been neutralized, requiring 0.0100 mol NaOH. The amount of H2PO4 – and HPO4 2- is given by the Henderson-Hasselbalch equation:
6.50 = 7.21 + log([HPO4 2-]/[H2PO4 –]), so [HPO4 2-]/[H2PO4 –] = 0.195. The total phosphate concentration is 0.100 M, so [HPO4 2-] = (0.195/1.195)•0.100 M = 0.0163 M, requiring an additional 0.00163 mol NaOH to achieve. Thus a total of 0.0116 mol of NaOH must be added.
pH = 13.00
Model Answer
As in (i), the first two protons must be neutralized (requiring 0.0200 mol NaOH), then:
13.00 = 12.32 + log([PO4 3-]/[HPO4 2-]), so [PO4 3-]/[HPO4 2-] = 4.79
[PO4 3-] = 0.100 M•(4.79/5.79) = 0.0827 M
This requires an additional 0.00827 mol NaOH to remove the third proton from H3PO4.
BUT WAIT! At pH = 13.00, [OH–] = 0.100 M. This must come from yet more NaOH, another 0.0100 mol of it. (This was not an issue at pH = 6.5, where the OH– concentration was much smaller than the concentrations of the phosphate ions and so made a negligible difference in the amount of hydroxide one needed to add.) So the total amount of NaOH one must add is 0.0200 mol + 0.00827 mol + 0.0100 mol = 0.0383 mol NaOH.
An alternative approach to this problem would involve charge balance considerations.
The anions present in significant concentration at pH = 13 are HPO4 2- (0.0173 M), PO4 3– (0.0827 M), and OH– (0.100 M). The only cation present in significant concentration is Na+, so [Na+] = 2[HPO4 2–] + 3[PO4 3–] + [OH–] = 2[0.0173] + 3[0.0827] + [0.100] = 0.383 M. It would require adding 0.0383 mol NaOH to 100 mL to achieve this concentration of Na+.
Explain why one cannot prepare an effective buffer solution at pH = 10.0 using only 0.100 M H3PO4 and solid NaOH.
Model Answer
At pH = 10.0, almost all the phosphate must be in the form of HPO4 2-, with less than 1% in the form of either H2PO4 2- or PO4 3-. So any addition of acid will result in a significant increase of [H2PO4 –] and hence a significant drop in pH, and addition of base will result in a significant increase in [PO4 3-] and a significant rise in pH. Without a significant concentration of both an acid and its conjugate base, buffering is ineffective.