25.0 cm3 of an acidified solution containing Fe2+ ions was titrated against potassium manganate (VII) solution. 20.0cm3 of 0.050M potassium manganate (VII) was needed. Calculate the concentration of Fe2+ ions in the acidified solution.

Write the equation and balance it.

(If you mean potassium manganate that is K2MnO4. If you mean potassium permanganate that is KMnO4. I suspect you mean KMnO4 since yu write (VII).
Using the coefficients in the balanced equation, convert moles KMnO4 to mols Fe^+2. Then M Fe = moles Fe/L Fe.

.000025MFe ???

To determine the concentration of Fe2+ ions in the acidified solution, we can use the balanced equation for the reaction between Fe2+ ions and potassium manganate (VII):

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O

From the equation, we can see that 5 moles of Fe2+ ions react with 1 mole of MnO4- ion. Therefore, the ratio of Fe2+ to MnO4- is 5:1.

Since 20.0 cm3 of 0.050 M potassium manganate (VII) solution is needed to react with the Fe2+ ions, we can determine the number of moles of MnO4- used in the titration:

Volume of potassium manganate (VII) solution = 20.0 cm3 = 20.0 mL
Molarity of potassium manganate (VII) solution = 0.050 M

Number of moles of MnO4- used = Molarity × Volume
= 0.050 M × 20.0 mL
= 0.00100 moles

Since the ratio of Fe2+ to MnO4- is 5:1, the number of moles of Fe2+ ions present in the solution is also 0.00100 moles.

Now, we can determine the concentration of Fe2+ ions in the solution:

Volume of acidified solution = 25.0 cm3 = 25.0 mL

Molarity of Fe2+ = Number of moles / Volume
= 0.00100 moles / 25.0 mL
= 0.0400 M

Therefore, the concentration of Fe2+ ions in the acidified solution is 0.0400 M.

To find the concentration of Fe2+ ions in the acidified solution, we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of the acidified solution (Fe2+ ions)
V1 = volume of the acidified solution (25.0 cm3)
M2 = concentration of potassium manganate (VII) solution (0.050 M)
V2 = volume of the potassium manganate (VII) solution used (20.0 cm3)

Rearranging the equation to solve for M1:

M1 = (M2 * V2) / V1

Plugging in the given values:

M1 = (0.050 M * 20.0 cm3) / 25.0 cm3

M1 = 0.04 M

Therefore, the concentration of Fe2+ ions in the acidified solution is 0.04 M.