Considering that the proton concentration corresponds to the analytic concentration of HNO3, [H+]aq = = 2.00 × mol we get [HA]org = 3.41 × Next, 10−2 dm−3, 10−3 mol dm−3. the uranyl ion distribution ratio, is expressed as: Using , , and i for [UO2(OH)i]2–i complexes, can be expressed as The concentration of hydroxyl ions is obtained from the concentration of protons, For [H+]aq = = 2.00 × mol we get 10−2 dm−3, mol dm−3 Casting this value, [HA]org = 3.41 × mol and all the necessary constants into 10−3 dm−3 the expression for the distribution ratio, we obtain Then, the yield R defined as can be calculated, providing the final result of 12.2 For the conditions of = = 2.00 × mol using the same calculation [OH−] 10−4 dm−3, procedure, we get [HA]org = 1.50 × mol 10−5 dm−3 and the yield R = 0.0121%. Problem 13. Determination of active chlorine in commercial products 13.1 (i) Cl2 + HClO (A) + HCl (B) H2O → (ii) NaClO + HClO (A) + NaOH (C) H2O → In alkaline aqueous solution, hypochlorite ion will dominate. (ClO−) 13.2 13.3 + 2 + 2 H+ I2 + H2O + ClO− I− → Cl− I2 + 2 2 + S2O32− → I− S4O62−
Considering that the proton concentration corresponds to the analytic concentration of HNO3, [H+]aq = = 2.00 × 10−2 mol dm−3, we get [HA]org = 3.41 × 10−3 mol dm−3. Next, the uranyl ion distribution ratio, is expressed as: Using , , and i for [UO2(OH)i]2–i complexes, can be expressed as The concentration of hydroxyl ions is obtained from the concentration of protons, For [H+]aq = = 2.00 × 10−2 mol dm−3, we get mol dm−3 Casting this value, [HA]org = 3.41 × 10−3 mol dm−3 and all the necessary constants into the expression for the distribution ratio, we obtain Then, the yield R defined as can be calculated, providing the final result of 12.2 For the conditions of [OH−] = = 2.00 × 10−4 mol dm−3, using the same calculation procedure, we get [HA]org = 1.50 × 10−5 mol dm−3 and the yield R = 0.0121%. Problem 13. Determination of active chlorine in commercial products 13.1 (i) Cl2 + H2O → HClO (A) + HCl (B) (ii) NaClO + H2O → HClO (A) + NaOH (C) In alkaline aqueous solution, hypochlorite ion (ClO−) will dominate. 13.2 13.3 ClO− + 2 I− + 2 H+ → I2 + H2O + Cl− I2 + 2 S2O32− → 2 I− + S4O62−
PrepProblems%2050%20IChO%20Theoretical%20Solutions.docx