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.