# Quantity of water in a redox problem

$\pu{76 g}$ of $\ce{FeSO4}$ are oxidized by $\pu{0.04 M}$ solution of $\ce{KMnO4}$ in presence of $80\%$ $\ce{H2SO4}$. What quantity of water must be added to the final solution of $\ce{Fe2(SO4)3}$ for the concentration to be $\pu{0.25 M}$?

I wrote the reaction:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

But I have no idea how to find the amount of water. I found the quantity of $\ce{Fe2(SO4)3}$ to be $\pu{100 g}$, but I have no idea to find all the water.

• Could you please double-check the amounts? I might be wrong, but something seems a bit off here. Probably the concentration of KMnO4 is 0.4 M, not 0.04 M? Dec 9 '17 at 8:44
• This is strange; in my opinion it's a typo, because this is a way too diluted solution to start with. Dec 13 '17 at 22:13

To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:

$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$

where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.

From the balanced redox reaction, which it seems you have established correctly:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:

$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$

$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$

$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$

From (1):

$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$

Combining (2–4) into (5), and taking into account that

$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$

one can obtain the following expression for the volume of water to be added:

$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$

The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:

$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$