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3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of water. 20.0 mL of the solution had some acid added to it and then it reacted completely with 12.6 mL of KMnO4 solution. Calculate the concentration of the KMnO4 solution given the full REDOX equation below.

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

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Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

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

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

[tex]1.4 * 10^{-4 }M[/tex]

The balanced redox reaction equation is shown below:

[tex]5Fe^{2+} + MnO_4^{-} + 8H^{+}[/tex] → [tex]5Fe^{3+} +Mn^{2+} + 4H_2O[/tex]

Molar mass of FeSO₄(NH₄)₂SO₄*6H₂O = 392 g/mol  

Number of moles [tex]Fe^{2+}[/tex] in FeSO₄(NH₄)₂SO₄*6H₂O

[tex]\text{Number of Moles}= \frac{3.47g}{392g/mol} = 8.85 * 10^{-5} \text{ moles}[/tex]

Concentration of [tex]Fe^{2+}[/tex]  = [tex]8.85 * 10^{-5} \text{ moles} *\frac{1000}{200} = 4.425 *10^{-4} M[/tex]

Let [tex]C_A[/tex] be concentration of [tex]Fe^{2+}[/tex]  = [tex]4.425 * 10^{-4} M[/tex]  

Volume of [tex]4.425 * 10^{-4} M[/tex] ([tex]V_A[/tex])= 20.0 ml

Let the concentration of [tex]MnO_4^{-}[/tex]  be [tex]C_B[/tex] (the unknown)

Volume of the [tex]MnO_4^{-}[/tex] ([tex]V_B[/tex]) = 12.6 ml

Let the number of moles of [tex]Fe^{2+}[/tex]  be [tex]N_A[/tex]= 5 moles

Let the number of moles of[tex]MnO_4^{-}[/tex]  be [tex]N_B[/tex] = 1 mole

Typically for acid and base reactions, using a titration will help you determine how much acid or base is needed to neutralize the other, and also the concentration and molarity of the solution  which is given as"  

[tex]\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B} \\\\C_AV_AN_B = C_BV_BN_A\\\\C_B= C_AV_AN_B/V_BN_A\\\\C_B= 4.425 * 10^{-4} * 20 * 1/12.6 * 5\\\\C_B = 1.4 * 10^{-4} M[/tex]

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