Question

Can I get all of the question answered in a way that I know under which sub question the answer goes please.

Can I get all of the questions answered in a way that I know how it fits in the question. Some parts are explanations. Thank you.

Q. When a pink aqueous solution of potassium permanganate, faintly acidified with dilute sulfuric acid was treated with 10% aq. hydrogen peroxide, the reaction took place with the evolution of gas bubbles, and the pink solution was turned colorless. Further chemical analysis revealed that the evolved gas was oxygen, and the resulting solution contains potassium sulfate and manganese (II) sulfate; water was also formed during the same reaction. Please answer the followings:

1) Write down the balanced chemical equation for this reaction.

2) Define the type/types of the reaction; assign the oxidation number of Manganese in potassium permanganate and manganese (II) sulfate. Also, offer an explanation for the color change.

3) Write down the ions present in the solution before & after the reaction. If 100 mL 0.5 M potassium permanganate was mixed with 50 mL 2 M sulfuric acid, what will be the final pH (>7 or <7) of the solution?

4) After the completion of reaction, when evolution of gas-bubbles ceased, what will happen if an aqueous solution of sodium carbonate was added to the reaction mixture? Please write down the net ionic equation.

5) If initially 5.65 g of potassium permanganate was taken for the reaction, calculate the total mass of manganese (II) sulfate present in the solution after the completion of the reaction. What is the Manganese ion concentration of the solution (assume that the total volume of the final reaction mixture is 500 mL)?

Useful hints: hydrogen peroxide is a strong oxidizing agent and considered to be a molecular compound.

Answer 1

1) Write down the reaction first:

KMnO₄ + H₂O₂ + H₂SO₄ ------> K₂SO₄ + MnSO₄ + O₂ + H₂O

To balance the equation, we need to assign the oxidation numbers first.

Mn in KMnO₄ = +7

Mn in MnSO₄ = +2

O in H₂O₂ = -1

O in O₂ = 0

Change in oxidation number per atom of Mn = (+7) – (+2) = 5

Change in oxidation number per atom of O =(0) – (-1) = 1

Since we have two atoms of O in H₂O₂, the change in oxidation number for O = (1)*2 = 2

To balance the reaction, cross-multiply; this gives

2 KMnO₄ + 5 H₂O₂ + H₂SO₄ ------> K₂SO₄ + 2 MnSO₄ + 5 O₂ + H₂O

Finally balance the chemical equation by adjusting the co-efficients for H₂SO₄ and H₂O. The balanced chemical equation is

2 KMnO₄ + 5 H₂O₂ + 3 H₂SO₄ -----> K₂SO₄ + 2 MnSO₄ + 5 O₂ + 8 H₂O

2) The reaction is a redox reaction where Mn(VII) in KMnO₄ is reduced to Mn(II) in MnSO₄ and H₂O₂ (oxidation number of O in H₂O₂ is -1) is oxidized to O₂ (oxidation number of O is 0).

The colour change occurs due to the reduction of Mn(VII) to Mn(II). Mn(VII) is intense pink in aqueous solution due to increased electronic transitions (has more available orbitals) while Mn(II) is pale in colour due to fewer transitions.

3) H₂O₂ is a molecular compound and doesn’t dissociate into ions. Similar is the case with molecular oxygen, O₂. KMnO₄ is a strong electrolyte and dissociates completely while both K₂SO₄ and MnSO₄ dissociate into ions completely. The ions present in the solution before the reaction is

K⁺, MnO₄^-, H⁺, SO₄^2-

The ions present in the solution after the reaction are: K⁺, Mn²⁺, SO₄^2-

Since the reaction takes place in 2 M sulfuric acid solution, the pH of the solution before the reaction will be less than 7 and the medium is acidic.

4) After the reaction is complete, KMnO₄ and H₂O₂ are completely neutralized to MnSO₄ and O₂. Neither K₂SO₄, MnSO₄ or O₂ are acidic in nature and hence will not react with sodium carbonate, Na₂CO₃, which is basic in solution. The solution contains H₂SO₄ which can react with Na₂CO₃ to evolve CO_2.

Na₂CO₃ + H₂SO₄ -----> Na₂SO₄ + CO₂ + H₂O

The ionic equation is

2 Na⁺ + CO₃^2- + 2 H⁺ + SO₄^2- ----> 2 Na⁺ + SO₄^2- + CO₂ + H₂O

Cancel out the common ions to obtain the net ionic equation as

CO₃^2- + 2 H⁺ -----> CO₂ + H₂O

5) Look at the balanced chemical equation above; the molar ratio of KMnO₄ and MnSO₄ is 1:1, i.e.,

1 mole KMnO₄ = 1 mole Mn(II)

We have 5.65 g KMnO₄; molar mass of KMnO₄ = 158.034 g/mol.

Therefore, moles of KMnO₄ taken = (5.65 g)/(158.034 g/mol) = 0.03575 mol.

Moles of MnSO₄ = (0.03575 mol KMnO₄)*(1 mol MnSO₄/1 mol KMnO₄) = 0.03575 mol.

Molar mass of MnSO₄ = 151.001 g/mol.

Mass of MnSO₄ obtained = (0.03575 mol)*(151.001 g/1 mol) = 5.398 (ans).

Concentration of Mn(II) = moles of MnSO₄/volume of solution in L = (0.03575 mol)/(0.500 L) = 0.0715 mol/L = 0.0715 M (ans).