# An iron nail with a mass of 1.365 g is soaked in an acidic solution, yielding Fe2+ solution, which is then titrated with 26.48 mL of KMnO4 solution in the following equation: Calculate the molar concentration of the KMnO4 solution?

## Fe2+ + MnO4- → Mn2+ + Fe3+

May 30, 2018

$\left[M n {O}_{4}^{-}\right] = 0.18 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We gots a series of redox reactions...

$F e \left(s\right) + 2 H C l \left(a q\right) \rightarrow F {e}^{2 +} + {H}_{2} \left(g\right) \uparrow + 2 C {l}^{-}$

And then ferrous ion is OXIDIZED up to ferric ion by the action of permanganate:

$F {e}^{2 +} \rightarrow F {e}^{3 +} + {e}^{-}$ $\left(i\right)$

${\underbrace{M n {O}_{4}^{-}}}_{\text{strongly coloured}} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O \left(l\right)$ $\left(i i\right)$

And we add $5 \times \left(i\right) + \left(i i\right)$

$5 F {e}^{2 +} + M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow 5 F {e}^{3 +} + M {n}^{2 +} + 4 {H}_{2} O \left(l\right) + 5 {e}^{-}$

...and on cancellation we get the stoichiometric equation...

$5 F {e}^{2 +} + M n {O}_{4}^{-} + 8 {H}^{+} \rightarrow 5 F {e}^{3 +} + {\underbrace{M {n}^{2 +}}}_{\text{almost colourless}} + 4 {H}_{2} O \left(l\right)$

The colour change of the manganese ion allows determination of a good stoichiometric endpoint.

And now we calculate the molar equivalence...

$\text{Moles of iron} = \frac{1.365 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1} = 0.0244 \cdot m o l$..but CLEARLY, from the given stoichiometry, there were $\frac{1}{5} \cdot \text{equiv}$ with respect to permanganate...

And so [MnO_4^-]=(0.0244*molxx1/5)/(26.48xx10^-3*L)=??*mol*L^-1

May 30, 2018

["KMnO"_4]=0.1846 color(white)(l) "M"

#### Explanation:

Steps:

1. Calculate $n \left({\text{Fe}}^{2 +}\right)$ from mass;
2. Derive the value of $n \left({\text{MnO}}_{4}^{-}\right)$ concerning stoichiometrical relationships in redox reaction;
3. Calculate the molarity of potassium permangnate $\left[{\text{KMnO}}_{4}\right]$ with the equation $c = \frac{n}{V}$.

Assume that

• the entire $1.365 \textcolor{w h i t e}{l} \text{g}$ nail contains no ${\text{Fe}}^{3 +}$, the oxidation product of the titration process, and

• the nail dissolves completely in the acid.

The number of iron atoms conserves. Molar mass of iron M("Fe")=55.85 color(white)(l)"g" * "mol"^(-1);

n("Fe"^(2+))=n("Fe")=(m("Fe"))/(M("Fe"))=2.444 xx 10^(-2) color(white)(l) "mol".

The reaction formula given in the question isn't properly balanced. However, given that changes in oxidation state cancel out in the net reaction, it is possible to obtain the coefficient ratio straight from changes in oxidation states without having to balance the whole equation. Start by identifying the oxidation state in each of the species:

Reactants:

• $\stackrel{\boldsymbol{\textcolor{n a v y}{+ 2}}}{\text{Fe}} {\textcolor{w h i t e}{.}}^{2 +}$
• ${\text{K"stackrel(bb(color(purple)(+7)))("Mn")"O}}_{4} {\textcolor{w h i t e}{.}}^{-}$

Products:

• $\stackrel{\boldsymbol{\textcolor{n a v y}{+ 3}}}{\text{Fe}} {\textcolor{w h i t e}{.}}^{3 +}$
• $\stackrel{\boldsymbol{\textcolor{p u r p \le}{+ 2}}}{\text{Mn}} {\textcolor{w h i t e}{.}}^{2 +}$

Iron $\text{Fe}$ is oxidized; its oxidation state increased by $\textcolor{n a v y}{\boldsymbol{1}}$, from $+ 2$ to $+ 3$;
Manganese is reduced; its oxidation state decreased by $\textcolor{p u r p \le}{\boldsymbol{5}}$, from $+ 7$ to $+ 2$.

Sum of rises in the oxidation state $=$ sum of drops in the oxidation state; Iron and manganese are the only two elements with oxidation state changes in this particular reaction; therefore

$\textcolor{n a v y}{1} \cdot n \left(\stackrel{\boldsymbol{\textcolor{n a v y}{+ 2}}}{\text{Fe"))=color(purple)(5)*n(stackrel(bb(color(purple)(+7)))("Mn}}\right)$

• n("Fe"^(2+))=n(stackrel(bb(color(navy)(+2)))("Fe"))
• n("KMnO"_4)=n("MnO"_4^(-))=n(stackrel(bb(color(purple)(+7)))("Mn"))

Hence

n("KMnO"_4)=color(navy)(1)/color(purple)(5)*n("Fe"^(2+))=4.888 xx 10^(-3) color(white)(l) "mol"

The question states that the $4.888 \times {10}^{- 3} \textcolor{w h i t e}{l} \text{mol}$ of ${\text{KMnO}}_{4}$ is dissolved in 26.48 color(white)(l) "ml"=2.648 xx 10^(-2) color(white)(l) color(green)("L") of solution. Note that the unit for molarity, $\textcolor{\mathrm{da} r k g r e e n}{\text{M}}$ is defined as "mol" * color(green)("L")^(-1) or equivalently "mol" * color(green)("dm")^(-3), meaning that the volume of the solution shall be represented in $\text{L}$ in molarity calculations.

Therefore

["KMnO"_4]=(n("KMnO"_4))/(V("KMnO"_4))
$\textcolor{w h i t e}{\left[\text{KMnO"_4])=(4.888 xx 10^(-3) color(white)(l) "mol")/(2.648 xx 10^(-2) color(white)(l) color(green)("L}\right)}$

color(white)(["KMnO"_4])=0.1846 color(white)(l) "mol" * color(green)("L")^(-1)
color(white)(["KMnO"_4])=0.1846 color(white)(l) color(darkgreen)("M").