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- Sulfuric acid (H2SO4) is called a diprotic acid because it has two acidic protons. The pKa for the first deprotonation is -9, whereas the pKa for the second deprotonation is 2. Explain these relative acid strengths.Two dyes (orange, MW = 300 and blue, MW = 100,000) are placed indialysis tubes and put into beakers of water. Explain why there isorange dye and not blue dye in the beaker waterIf an Ecell of 0.495V was noted when your cell was set-up. What is thesolution pH?
- The following cell was found to have a potential of —0.492 V: Ag|AgCl(sat’d)||HA(0.200 M),NaA(0.300 M)|H2(1.00 atm),Pt Calculate the dissociation constant of HA, neglecting the junction potential.Acetic acid (CH3COOH, abbrev HA) is a very common weak acid with pKa = 4.75 (at 25oC and low ionic strength) A] Write down the dissociation equation of the weak acid HA in aqueous solution, derive the respective formulae for the dissociation constant Ka and pKa. B] Calculate pH of the diluted solution of acetic acid with concentration, c = 0.05 mol/dm3 C] What is the approximate pH of the aqueous acetate buffer solution composed of dissolved mixture of acetic acid (HA) and sodium acetate (NaA) in the same respective concentrations? What is the working pH range of acetate buffers? D] How can you prepare exactly 1 dm3 of acetate buffer solution with required pH 5.0 and the total concentration [HA + NaA], c = 0.01 mol/dm3 if the stock solution of acetic acid (HA) with concentration, cHA = 0.1 mol/dm3, anhydrous sodium acetate (NaA, MT = 82.03 g/mol) and purified water are available? E] Acetate buffer solution may be also prepared by the partial neutralization of HA with NaOH: 20 cm3 of…Question 2. (A) Choose one ionisation technique from the ones discussed in the course (EI, CI, MALDI, ESI) and provide a detailed explanation of how it works, including diagrams if you like. Cover typical applications, and pros/cons, mass analysers that are typically used with this source etc.
- Dihydrogen phosphate, H₂PO₄⁻(aq), undergoes the following acid base equilibria in blood: H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq)⇌ H⁺(aq) + PO₄³⁻(aq). The pKa values for the first and second steps are 7.2 and 12.4, respectively. The pH of human blood is 7.37 and this is tightly regulated. Which of the following statements most accurately describes the concentrations of the different species for the above reactions in blood. (Hint, think of the titration curves for H₂PO₄⁻(aq) and HPO₄²⁻(aq) and their respective pKa's.) A) [H₂PO₄⁻(aq)] is much greater than [HPO₄²⁻(aq)] and [PO₄³⁻(aq)] B) [HPO₄²⁻(aq)] is much greater than [H₂PO₄⁻(aq)] and [PO₄³⁻(aq)]. C) [H₂PO₄⁻(aq)] and [HPO₄²⁻(aq)] are both present and greater than [PO₄³⁻(aq)]. D) [PO₄³⁻(aq)] is much larger than the other concentrations, [H₂PO₄⁻(aq)] and [HPO₄²⁻(aq)]Calculate (to the nearest 0.1%) the proportion of a dose of ebastine (pKa of conjugate acid = 10.3) that will be ionised at pH 7.7. You do not need to include "%" in your answer.Assign the following as Ror S.
- (9.29 × 105/8)-20.81 = with correct sig figsspectrochemical series which is the interaction relationship of Lewis bases with metals oftransition: a) Explain the order of each base in relation to Fe3+. Why?b) Determine the order of bases with respect to NH4+. Why?2 Two nitroimidazole drugs, misonidazole and tinidazole have single-electron reduction potentials of -0.389 V and -0.464 V respectively: Misonidazole + e- ➝ Misonidazole·- E0 = -0.389 V Tinidazole + e- ➝ Tinidazole·- E0 = -0.464 V The reduction potential of the NADH/NAD couple is -0.324 V, and the reduction potential of the ferrodoxin couple is -0.460 V: NAD+ + e- ➝ NADH E0 = -0.324 V [Ferrodoxin]3+ + e- ➝ [Ferrodoxin]2+ E0 = -0.460 V Explain which, if any, of misonidazole and tinidazole will be selectively reduced in protozoa over mammalian cells. In your answer you should calculate any reduction potentials you need.