±Titrations and Solution Stoichiometry 9 of 18> Constants I Periodic Table Part
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Question
±Titrations and Solution Stoichiometry 9 of 18> Constants I Periodic Table Part A A titration is a procedure for determining the concentration of a solution by allowing it to react with another solution of known concentration (callied a standard solution). Acid-base reactions and oxidation- reduction reactions are used in titrations. For example, to find the concentration of an HCl solution (an acid), a standard solution of NaOH (a base) is added to a measured volume of HCI from a calibrated tube called a buret. An indicator is also present and it will change color when all the acid has reacted. Using the concentration of the standard solution and the volume dispensed, we can calculate molarity of the HCI solution. A volume of 50.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2 the molarity of the KOH solution if 18.2 mL of 1.50 M H2SO4 was needed? The equation is SO4). What was 2KOH(a)+H2S04(aq)+K2SO4(aq) +2H20(1) Express your answer with the appropriate units. View Available Hint(s) molarityValue Units Submit Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H202, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMn04 (a)+ H202(ag) +3H2SO4(aq) 30, (g) + 2MnSO4(aq) + K So4(aq) +4H204) n peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4 What mass of H202 was dissolved if the titration required 19.3 mL of the KMnO4 solution? Express your answer with the appropriate units. View Available Hint(s)Explanation / Answer
Part a
Moles of H2SO4 = molarity x volume
= 1.50 mol/L x 18.2 mL x 1L/1000 mL
= 0.0273 mol
From the stoichiometry of the reaction
Moles of KOH required
= 0.0273 mol H2SO4 x 2 mol KOH/1 mol H2SO4
= 0.0546 mol KOH
Molarity of KOH solution
= moles of KOH / volume of KOH solution
= (0.0546 mol)/ (50 mL x 1L/1000 mL)
= 1.092 M
Part b
Moles of KMnO4 = molarity x volume
= 1.68 mol/L x 19.3 mL x 1L/1000 mL
= 0.032424 mol
From the stoichiometry of the reaction
Moles of H2O2 required
= 1 mol H2O2 x 0.032424 mol KMnO4 / 2 mol KMnO4
= 0.016212 mol H2O2
Mass of H2O2 = moles x molecular weight
= 0.016212 mol x 34.01 g/mol
= 0.551 g
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