A titration is a procedure for determining the concentration of a solution by al

Question

A titration is a procedure for determining the concentration of a solution by allowing it to react with another solution of known concentration (called a standard solution). Acid-base reactions and oxidation-reduction reactions are used in titrations. For example, to find the concentration of an HC1 solution (an acid), a standard solution of NaOH (a base) is added to a measured volume of HCl 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 HCl solution. A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H_2SO_4). What was the molarity of the KOH solution if 20.7 mL of 1.50 M H_2SO_4 was needed ? The equation is 2KOH(aq) + H_2SO_4(aq) rightarrow K_2SO_2 (aq) + 2H_2O(1) Express your answer with the appropriate units. Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H_2O_2, can be titrated against a solution of potassium permanganate, KMnO_4. The following equation represents the reaction: A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO_4. What mass of H_2 O_2 was dissolved if the titration required 20.8 mL of the KMnO_4 solution? Express your answer with the appropriate units.

Explanation / Answer

Part A)

We see from the equation that 2 moles of KOH completely reacts with 1 mole of H2SO4

Given,

Volume of KOH solution = 70 mL = 0.07 L

Volume of H2SO4 solution = 20.7 mL = 0.0207 mL

Molarity of H2SO4 = 1.5 M

Moles = Molarity x Volume (in L)

=> Moles of H2SO4 in solution = 1.5 x 0.0207 = 0.03105

=> Moles of KOH required = 0.03105 x 2 = 0.0621

Molarity = Moles / Volume (L)

=> Molarity of KOH = 0.0621 / 0.07 = 0.887 M

2) According to the reaction, 2 moles of KMnO4 reacts with 1 mole of H2O2

Given,

Volume of KMnO4 solution = 20.8 mL = 0.0208 L

Molarity of KMnO4 = 1.68 M

Volume of H2O2 solution = 100 mL = 0.1 L

Moles of KMnO4 = 1.68 x 0.0208 = 0.034944 moles

=> Moles of H2O2 required = 0.034944 / 2 = 0.017472 moles

Molar mass of H2O2 = 34.0147 g / mol

=> Mass of H2O2 = 0.017472 x 34.0147 = 0.594 g

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