# In order to standardize a KMnO4 solution, 0.2848 g Fe(NH4)2(SO4)2·6H2O was dissolved in 25 mL 0.18 M H2SO4. The KMnO4 solution was added to the Fe(NH4)2(SO4)2·6H2O solution until a pale pink color persisted. The titration took 24.2 mL of KMnO4 solution. What is the concentration of the KMnO4 solution?

## No, no, Bridget. That will not work in this case BECAUSE 1 mol Fe is not = 1 mol MnO4^-.

1. Write and balance the equation.

Fe(NH4)2(SO4)2.6H2O + KMnO4 + H2SO4 ==> Fe2(SO4)3 + (NH4)2SO4 + K2SO4 + MnSO4 + H2O

2. Convert 0.2848g Fe(NH4)2(SO4)2.6H2O to mols. mol = grams/molar mass = ?

3. Using the coefficients in the balanced equation, convert mols Fe(NH4)2(SO4)2.6H2O to mols KMnO4.

4. Then M KMnO4 = mols KMnO4/L KMnO4. You know mols and L, solve for M.

## You have to use the M1V1=M2V2 equation for this problem and plug in the values into the appropriate parts of the equation and solve for the missing molar concentration.

## To calculate the concentration of the KMnO4 solution, we can use the equation:

M1V1 = M2V2

Where:

M1 - concentration of KMnO4 solution

V1 - volume of KMnO4 solution used in the titration

M2 - concentration of Fe(NH4)2(SO4)2·6H2O solution

V2 - volume of Fe(NH4)2(SO4)2·6H2O solution

Given:

M2 = 0.18 M

V2 = 25 mL

V1 = 24.2 mL

Rearranging the equation, we have:

M1 = (M2 * V2) / V1

Substituting the given values, we get:

M1 = (0.18 M * 25 mL) / 24.2 mL

Calculating this equation, we find:

M1 = 0.185 M

Therefore, the concentration of the KMnO4 solution is 0.185 M.

## To determine the concentration of the KMnO4 solution, we can use the concept of stoichiometry and the known reaction between Fe(NH4)2(SO4)2·6H2O and KMnO4.

The balanced equation for the reaction is:

5Fe(NH4)2(SO4)2·6H2O + 8KMnO4 + 34H2SO4 → 10FeSO4 + 5H2SO4 + 8MnSO4 + K2SO4 + 14NH4HSO4 + 6H2O

From the reaction, we can see that the ratio of Fe(NH4)2(SO4)2·6H2O to KMnO4 is 5:8. Therefore, we can set up the following equation using the given mass of Fe(NH4)2(SO4)2·6H2O and the volume of KMnO4 used in the titration:

(0.2848 g Fe(NH4)2(SO4)2·6H2O / molar mass of Fe(NH4)2(SO4)2·6H2O) / (24.2 mL KMnO4 / 1000 mL) = (n moles KMnO4 / 8)

First, let's calculate the moles of Fe(NH4)2(SO4)2·6H2O:

(0.2848 g Fe(NH4)2(SO4)2·6H2O / molar mass of Fe(NH4)2(SO4)2·6H2O)

= (0.2848 g / 392.14 g/mol)

≈ 0.0007273 moles

Next, we rearrange the equation and find the moles of KMnO4 used in the titration:

(n moles KMnO4) = (0.0007273 moles Fe(NH4)2(SO4)2·6H2O) * (8 / 24.2)

≈ 0.0002399 moles

Finally, to find the concentration of KMnO4, we divide the moles of KMnO4 by the volume of KMnO4 used in the titration:

(concentration of KMnO4) = (0.0002399 moles / 0.0242 L)

≈ 0.009918 M

Therefore, the concentration of the KMnO4 solution is approximately 0.009918 M.