Asked by tam
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.
Answers
Answered by
DrBob222
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.
(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.
Answered by
Jillian
.000025MFe ???