Question

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant. They titrated a 12.50 mL aliquot of the Fe+2 solution with 0.0215 M MnO4- solution and determined that the equivalence point was at 10.8 mL. What is the molarity of the Fe+2 solution?

Answer 1

Answer:

Molarity of $Fe^{2+}$ solution is 0.0928 M.

Explanation:

Balanced equation: $5Fe^{2+}+MnO_{4}^{-}+8H^{+}\rightarrow 5Fe^{3+}+Mn^{2+}+4H_{2}O$

Number of moles of $MnO_{4}^{-}$ in 10.8 mL of 0.0215 M $MnO_{4}^{-}$ solution

= $\frac{0.0215}{1000}\times 10.8moles$ = 0.000232 moles

Let's assume molarity of  $Fe^{2+}$ solution is S(M)

Number of moles of $Fe^{2+}$ in 12.50 mL of S(M) $Fe^{2+}$ solution

= $\frac{S}{1000}\times 12.50moles$ = 0.0125S moles

According to balanced equation, 5 moles of  $Fe^{2+}$ react with 1 mol of $MnO_{4}^{-}$

So, 0.0125S moles of $Fe^{2+}$ react with $\frac{0.0125S}{5}$ moles of $MnO_{4}^{-}$

Hence, $\frac{0.0125S}{5}=0.000232$

         or, S = 0.0928

So, molarity of $Fe^{2+}$ solution is 0.0928 M.