A volume of 50.0 mL of aqueous potassium hydroxide (KOH) was titrated against a

Question

A volume of 50.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2 SO4). What was the molarity of the KOH solution if 13.7 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq) + H2SO4 (aq)K2SO4 (aq) + 2H20(1) Express your answer with the appropriate units. Hints 2 molarity = 1 Value Units Submit My Answers Give Up Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2 O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4 (aq) + H2O2 (aq) .+ 3H2 SO4 (aq) 302(g) 2MnSO4 (aq)K2SO4(a)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 H202 was dissolved if the titration required 15.8 mL of the KMnO solution? Express your answer with the appropriate units. Hints

Explanation / Answer

ANswer:

Part A:

Given equation 2KOH+H2SO4------> K2SO4+2H2O

Molarity of KOH (M1)=?, Volume (V1)= 50 mL, Molarity of H2SO4 (M2)=1.5 M, Volume of H2SO4 (V2)=13.7 mL

We know that M1V1=M2V2

M1=M2V2/V1=(1.5 Mx13.7 mL)/50 mL

M1=0.411 M.

The molarity of KOH =0.411 M.

Part B:

Given Molarity and volume of KMnO4 are M1=1.68 M, V1=15.8 mL,

And for H2O2, M2=? V2=100 mL

By using same formula M1V1=M2V2

1.68 Mx15.8 mL=M2x100 mL

M2=0.265 M

Molarity= (mass/molar mass)x1000/V(mL)

Molar mass of H2O2=34.014 g/mol, V=100 mL

0.265 mol/L=(mass/34.014 g/mol)x(1000/100 L)

mass=0.902 g

Therefore, 0.902 g of H2O2 required.

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