Title DCC Part II - Solutions.pdf ✅

DChO Chemistry Competition 2024 − Part II National Mock Written by Dr. Chen and his team 1 Q1. [14%] An iodate mineral has the chemical formula of M(IO3)Zy, where M is a metal cation, and Z is an anion. The ratio of the three ions in the mineral is 1:1:y. The molar mass of the mineral is 255.46 g mol-1 . a. When 1.000 g of the mineral was heated gently, a vapor was produced with a density of 0.5588 g/L at 120.0 °C and 1.000 atm. The mass of the residue was measured to be 0.9647 g. What is the molar mass of the vapor? What is the possible identity of Z? According to the ideal gas law, pV = nRT, we can further derive d = M×(p/RT) 1 pt Thus, M = dRT/p = 0.5588×0.08206×(120.0+273.2)/1.000 = 18.03 g/mol 1 pt It equals the molar mass of H2O vapor. Thus, Z = OH− . 1 pt b. 1.000 g of the mineral was dissolved in 30 mL of 2 mol/L HNO3. The resulting solution was made up of 100.0 mL with distilled water, forming a stock solution. 10.00 mL of this stock solution reacts with excess potassium iodide. The iodine formed was titrated with 0.1000 mol/L sodium thiosulfate solution. Assume that all iodine was formed in the reaction of iodate ions with iodide ions. Calculate the volume of sodium thiosulfate solution needed to react with the iodine produced. IO3 − (aq) + 5 I− (aq) + 6 H+ (aq) = 3 I2(aq) + 3 H2O(l) or IO3 − (aq) + 8 I− (aq) + 6 H+ (aq) = 3 I3 − (aq) + 3 H2O(l) I2(aq) + 2 S2O3 2− (aq) = 2 I − (aq) + S4O6 2− (aq) or I3 − (aq) + 2 S2O3 2− (aq) = 3 I− (aq) + S4O6 2− (aq) Equations above are not required if the correct stoichiometry is shown. n(IO3 − ) in 10.00 mL = 1.000/255.46×10.00/100.0 = 3.915×10-4 (mol) 1 pt n(I2 or I3 − ) = 3×n(IO3 − ) = 3×3.915×10-4 = 1.174×10-3 mol 1 pt n(S2O3 2− ) = 2× n(I2 or I3 − ) = 2×1.174×10-3 = 2.349×10-3 mol 1 pt V(S2O3 2− ) = 2.349×10-3 /0.1000 = 23.49 mL 1 pt c. However, the metal ion Mn+ in this mineral also reacts with iodide ions to form additional iodine. Therefore, the actual volume of sodium thiosulfate solution required in the titration was found to be 27.40 mL. Calculate the number of moles of iodine produced by the reaction of one mole of metal ion Mn+ with excess iodide ions. n(I2 or I3 − ) produced by Mn+ = (27.40 − 23.49)×0.1000×1/2 = 1.955×10-4 mol 1 pt n(Mn+ ) = n(IO3 − ) in 10.00 mL = 3.915×10-4 mol; n(I2 or I3 − ): n(Mn+ ) = 1.955×10-4 :3.915×10-4 ≈ 1:2 1/2 mol I2 per 1 mol of Mn+ 1 pt
DChO Chemistry Competition 2024 − Part II National Mock Written by Dr. Chen and his team 2 d. Deduct the formula of the mineral by reasonable analysis with computation. If y = 1, M(Mn+ ) = 255.46 − 17.01 − 174.90 = 63.55, M = Cu; 1 pt If y = 2, M(Mn+ ) = 255.46 − 2×17.01 − 174.90 = 46.54, no metals match; If y = 3, M(Mn+ ) = 255.46 − 3×17.01 − 174.90 = 29.53, no metals match; If y = 4, M(Mn+ ) = 255.46 − 4×17.01 − 174.90 = 12.52, no metals match; 1 pt The formula of the mineral is Cu(IO3)OH. 1 pt or confirm the value of y based on the mass of residue, shown below: n(H2O, g) = (1.000 − 0.9647)/18.02 = 1.96×10-3 mol = 1⁄2 n(mineral) à y = 1 e. Write down the balanced net ionic equation of Mn+ reacting with excess iodide. 2 Cu2+(aq) + 5 I − (aq, ex) = 2 CuI(s) + I3 − (aq) 2 pts full credit is provided if the product is I2 and the reaction is balanced. Note: According to part (c), each mole of Cu produces 0.5 mol of I2, thus, the oxidation state of Cu is reduced by 1, Cu2+ à Cu+ , however, Cu+ is unstable and CuI is a precipitate. 1 pt is provided if the product is written as Cu+ (aq) and the reaction is balanced. Q2. [12%] Histidine has the following structural formula: a. In an aqueous solution, histidine exists in the form of a zwitterion (electrically neutral) designated as HA. When the solution is acidified, histidine is protonated successively to form H2A+ and H3A2+. Draw the structural formula of the zwitterion form of HA. 2 pts No stereochemistry is needed. b. Calculate the pH of a 0.010 M HA solution. H3A2+: pKa1 = 1.82; pKa2 = 6.00; pKa3 = 9.17.
DChO Chemistry Competition 2024 − Part II National Mock Written by Dr. Chen and his team 3 HA is an amphoteric species, pH = (pKa2 + pKa3)/2 = (6.00 + 9.17)/2 = 7.59 2 pts or The following reaction occurs in the solution of HA: HA + HA ⇌ H2A+ + A− K = Ka3/Ka2 = 10-3.17 [H2A+ ] ≈ [A− ], and [HA] ≈ 0.010 M so K = [H2A+ ][A− ]/[A]2 = ([A− ]/[HA])2 = 10-3.17 [A− ]/[HA] = 10-3.17/2 = 10-1.59 HA ⇌ H+ + A− K = Ka3 = 10-9.17 Ka3 = [H+ ][A− ]/[HA] = 10-9.17 Substitute [A− ]/[HA] = 10-1.59 into the above expression: [H+ ] = 10-9.17+1.59 = 10-7.58 pH = 7.58 Note: Refer to the annotation of Q35 of Local 2022 or the solution to Q2b of National 2023 Part II for the derivation of the equation in the first method. c. Calculate the concentration of HClO4 required to establish a pH of 2.00 in a 0.010 M HA solution (solution X). pH = 2.00, close to pKa1 = 1.82, The major species in the solution are H3A2+ and H2A+ , According to the Henderson-Hasselbalch equation: pH = pKa1 + lg([H2A+ ]/[H3A2+]), [H2A+ ]/[H3A2+] = 102.00−1.82 = 1.5 1 pt [H2A+ ] + [H3A2+] = 0.010 M So, [H2A+ ] = 0.010×1.5/(1.5+1.0) = 6.0×10-3 M; [H3A2+] = 0.010×1/(1.5+1.0) = 4.0×10-3 M 1 pt Starting from HA, each H3A2+ needs two H+ , while H2A+ each need one H+ , So, c(HClO4) = 2×4.0×10-3 + 1×6.0×10-3 + 0.010 = 0.024 M 1 pt d. The complex formation of Cu2+ ions with histidine plays an important role in the interaction of this ion with proteins. Histidine can form the complexes CuA+ with Cu2+ ions. Mix 20. mL of solution X with 20. mL of a 0.010 M Cu2+ solution and dilute to 50. mL to obtain solution Y. Calculate the percentage of Cu2+ ions that form complexes with histidine in solution Y. Cu2+ + A− ⇌ CuA+ Kf = 1010.11 [Cu2+]0 = [H+ ]0 = 0.010×20./50. = 4.0×10-3 M [H3A2+]0 = 4.0×10-3 ×20./50. = 1.6×10-3 M [H2A+ ]0 = 6.0×10-3 ×20./50. = 2.4×10-3 M 1 pt
DChO Chemistry Competition 2024 − Part II National Mock Written by Dr. Chen and his team 4 Cu2+ + H3A2+ ⇌ CuA+ + 3 H+ K = Kf×Ka1×Ka2×Ka3 = 1010.11−1.82−6.00−9.17 = 10−6.88 Cu2+ + H2A+ ⇌ CuA+ + 2 H+ K = Kf×Ka2×Ka3 = 1010.11−6.00−9.17 = 10−5.06 Both Ks are very small, we can assume [CuA+ ] and free [A− ] are both negligible, The major species in the solution are still H3A2+ and H2A+ . 1 pt H3A2+ ⇌ H2A+ + H+ 1.6×10-3 −x 2.4×10-3 +x 4.0×10-3 +x (2.4×10-3 +x)×(4.0×10-3 +x)/(1.6×10-3 −x) = Ka1 = 10−1.82 x = 6.6×10-4 [H+ ] = 4.0×10-3 +x = 4.7×10-3 M [H2A+ ] = 2.4×10-3 +x = 3.1×10-3 M 1 pt H2A+ ⇌ 2 H+ + A− K = Ka2Ka3 [H+ ] 2 [A− ]/[H2A+ ] = (4.7×10-3 ) 2 ×[A− ]/(3.1×10-3 ) = Ka2Ka3 = 10−(6.00+9.17) [A− ] = 9.5×10-14 M 1 pt [Cu2+] ≈ [Cu2+]0 = 4.0×10-3 M Cu2+ + A− ⇌ CuA+ Kf = 1010.11 4.0×10-3 9.5×10-14 y y = 4.9×10-6 M CuA+ % = 4.9×10-6 /4.0×10-3 ×100% = 0.12% 1 pt Q3. [11%] Autoprotolysis also occurs in liquids other than water. For example, the autoprotolysis equilibrium constant of methanol is 10−16.70 at 25 °C. a. Write the chemical equation for the autoprotolysis of methanol. CH3OH(l) + CH3OH(l) ⇌ CH3OH2 + (solv) + CH3O− (solv) 2 pts b. Calculate the concentration of the cation of methanol under neutral conditions at 25 °C. [CH3OH2 + ] = [CH3O− ] = sqrt(10−16.70 ) = 4.5×10-9 M 2 pts c. 1-Ethylimidazolium bis(trifluoromethanesulfonyl)imide (hereafter denoted as CH+ /B− ), is an ionic liquid, an ionic compound that exists as a liquid even at room temperature.