Question

QUESTION 1: A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 12.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) QUESTION 2: Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)→3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 20.3 mL of the KMnO4 solution?

Answer 1

given: 2kom (aa) +4250 (aa) 4 (aa) - K2504 (aa) + K2504 202008) Volume of kon = som = 80x103 L volume of H₂SO4 = 12.2mL = 12.2010 Molasita g H2SO4 = 150M Moles g H₂SO4 = Volume of H2504x Molasity of H₂SO4 = (12. 2x10?y) * 150 mgeni = 0.0183 mon From ean.co, Molar ratio of kon to H₂504 18 2:1 Thus, Moles og kon= Moles of H2 504 x Molae ratio of kom to H2804 = 0.0183 mothy 504 X2 molkom mothes04 = 0.0366 mol kom Molarity of kon = Mcles og kon Volume og kon

= 0.0366 mol 80x 10²L -0.4575 molle = 0.458M @ gpwen: 2KMnO4 caa) + H20 (aa) KMNO4 (aq) +h2 caa) +312504 (aq) > 302 (9) + 2Mp Soacaq) + K2 504 (aa) 442060) --) Volume of KMnO4 = 203m2 = 20.3x1024 Molog?ta kuno4 - 1: 68м ' = 1.68 molle Males g KMnO4 = 20.3 xöxx1.68 mol 20.034104 mol from ean. (2), Molar rallo of H2O2 to KMnO4 Thus, Moles g H2O = Moles of KMnO4 x Molar ratio g H₂O2 to kina

= 0.034104 moh KMD04 Ximal 1202 amol Tomno = 0.017052 mol H2O2 Molar mass of M202 = 34.014791 mol Mass g 4202= Moles of H₂O2x Molar mass of H2O2 = 0.017052 rool x 34.01479 mot = 0.580018 6649 = 0.580g