A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a

Question

A volume of 80.0 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 19.2 mL of 1.50 M H_2SO_4 was needed? The equation is 2KOH (aq) + H_2SO_4 (aq) rightarrow K_2SO_4 (aq) + 2H_2O (l) Express your answer with the appropriate units. Molarity = 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: 2KMnO_4 (aq) + H_2O_2 (aq) + 3H_2SO_4 (aq) rightarrow 3O_2 (g) + 2MnSO_4 (aq) + K_2SO_4 (aq) + 4H_2O (l) 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_2O_2 was dissolved if the titration required 22.3 mL of the KMnO_4 solution? Express your answer with the appropriate units.

Explanation / Answer

mmol of acid = MV = 19.2*1.5 = 28.8 mmol

mmol of H+ = 2*28.8 = 57.6 mmol of H+

then

mmol of OH- = 57.6 mmol

[KOH] = [OH-] = mmol/mL = 57.6/80 = 0.72 M

2.

mmol of KMnO4 = 22.3*1.68 = 37.464 mmol of KMnO4

mmol of H2O2 = 37.464 /2 = 18.732 mmol of H2O2

then

[H2O2] = mmol/V= 18.732/100 = 0.18732 M

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