Question

In an oxidation-reduction reaction, it required 25.6 ml of a 0.65 M potassium permanganate solution to reach the equivalence point with 15.0 mL of an iron(II) sulfate solution. What is the molar concentration of the iron(II) sulfate solution? The net ionic equation for the reaction is:

Answer 1

Answer: The concentration of iron (II) sulfate solution is 5.66 M

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}` .....(1)

Molarity of potassium permanganate = 0.65 M

Volume of solution = 25.6 mL

Putting values in equation 1, we get:

`0.65M=\frac{\text{Moles of potassium permanganate}* 1000}{25.6}\\\\\text{Moles of potassium permanganate}=(0.65* 25.6)/(1000)=0.017mol`

Net ionic equation is defined as the equation in which no spectator ions are present in the reaction.

The net ionic equation for the reaction of potassium permanganate and iron (II) sulfate solution follows:

`MnO_4^-(aq.)+5Fe^(2+)(aq.)+8H^+(aq.)\rightarrow Mn^(2+)(aq.)+5Fe^(3+)(aq.)+4H_2O(l)`

By Stoichiometry of the reaction:

1 mole of permanganate ions react with 5 moles of iron (II) ions

So, 0.017 moles of permanganate ions will react with =
`(5)/(1)* 0.017=0.085mol` of iron (II) ions

Now, calculating the concentration of iron (II) ions by using equation 1, we get:

Moles of iron (II) sulfate = 0.085 mol

Volume of solution = 15.0 mL

Putting values in above equation, we get:

`\text{Molarity of }FeSO_4=(0.085* 1000)/(15.0)=5.66M`

Hence, the concentration of iron (II) sulfate solution is 5.66 M