Question

If a 0.12196 M solution of KMnO4 is titrated with a salt containing ferrous ion and 19.04 mL is required to reach the end point, calculate the moles of potassium permanganate used in the titration.

Give your answer to 4 places after the decimal. Do not enter units.

Answer 1

Question #72874, Chemistry / Other / Completed

If a 0.12196 M solution of $\mathrm{KMnO_4}$ is titrated with a salt containing ferrous ion and $19.04~\mathrm{mL}$ is required to reach the end point, calculate the moles of potassium permanganate used in the titration.

Solution

$$C_a = \frac{C_t V_t M}{V_a}$$

where $Ca$ is the concentration of the analyte, typically in molarity; $Ct$ is the concentration of the titrant, typically in molarity; $Vt$ is the volume of the titrant used, typically in liters; $M$ is the mole ratio of the analyte and reactant from the balanced chemical equation; and $Va$ is the volume of the analyte used, typically in liters.

$$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{Fe^{2+}} \longrightarrow 5\mathrm{Fe^{3+}} + \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$$

The quantity of $\mathrm{Fe^{2+}}$ ions are 5 times more than that of $\mathrm{KMnO_4}$ by the reaction. So $M = 1/5$

The moles of potassium permanganate:

$$n = C_a V_a = C_t V_t M = C_t \cdot 19.04~\mathrm{mL} \cdot \frac{1}{5} \cdot \frac{1}{1000} \cdot \frac{1}{1000}~\mathrm{means}~\mathrm{for}~1~\mathrm{liter}.$$ $$n = 0.003808~C_t$$

So if $C_t = 0.1~\mathrm{M}$, then $n = 0.0003808~\mathrm{mol}$.

**Answer: 0.003808 mol for $C_t = 0.1~\mathrm{M^*}$**

*We are not given enough of data that's why we use "if" and "for"!

The data table is attached:

**Answer provided by https://www.AssignmentExpert.com**