A titration is a procedure for determining the concentration of a solution by allowing it to...

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Chemistry

A titration is a procedure for determining theconcentration of a solution by allowing it to react with anothersolution of known concentration (called a standardsolution). Acid-base reactions and oxidation-reductionreactions are used in titrations. For example, to find theconcentration of an HCl solution (an acid), a standard solution ofNaOH (a base) is added to a measured volume of HCl from acalibrated tube called a buret. An indicator is alsopresent and it will change color when all the acid has reacted.Using the concentration of the standard solution and the volumedispensed, we can calculate molarity of the HCl solution.

Part A

A volume of 70.0 mL of aqueous potassium hydroxide (KOH) wastitrated against a standard solution of sulfuric acid (H2SO4). Whatwas the molarity of the KOH solution if 17.7 mL of 1.50 MH2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Express your answer with the appropriate units.

Redox titrations are used to determine the amounts of oxidizingand reducing agents in solution. For example, a solution ofhydrogen peroxide, H2O2, can be titrated against a solution ofpotassium permanganate, KMnO4. The following equation representsthe 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. mLof water and then titrated with 1.68 M KMnO4. What mass ofH2O2 was dissolved if the titration required 18.3 mL of the KMnO4solution?

Express your answer with the appropriate units.

Answer & Explanation Solved by verified expert
4.4 Ratings (704 Votes)

2KOH(aq) +   H2SO4(aq)→K2SO4(aq)+2H2O(l)

2 moles        1 moles

KOH                                                                     H2SO4

M1 =                                                                 M2 = 1.5M

V1 = 70ml                                                         V2 = 17.7ml

n1 = 2                                                                 n2 = 1

                  M1V1/n1     =     M2V2/n2

                         M1       = M2V2n1/n2V1

                                    = 1.5*17.7*2/1*70

                                    = 0.756M

B.

2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)→3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

2 moles          1mole

no of moles of KMnO4 = molarity * volume in L

                                      = 1.68*0.0183   = 0.03074moles

from balanced equation

2moles of KMnO4 react with 1 mole of H2O2

0.03074 moles of KMnO4 react with =1*0.03074/2 = 0.0154moles of H2O2

mass of H2O2 = no of moles * gram molar mass

                         = 0.0154*34 = 0.5236g >>> answer


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