A sample of pure tin metal is dissolved in nitric acid to produce 25.00 mL of a solution containing Sn^{2+}. When this tin solution is titrated, a total of 40.1 mL of 0.100 mol/L KMnO_{4} is required to reach the equivalence solution? Find the concentrat

The balanced chemical equation for the reaction between {eq}Sn^{2+} {/eq} and {eq}KMnO_{4} {/eq} is: {eq}5 Sn^{2+} + 2 MnO_{4}^{-} + 16 H^{+} rightarrow 5 Sn^{4+} + 2 Mn^{2+} + 8 H_{2}O {/eq} From the equation, we can see that each {eq}MnO_{4}^{-} {/eq} ion reacts with 5 {eq}Sn^{2+}{/eq} ions. Therefore, the number of moles of {eq}Sn^{2+} {/eq} in the solution can be calculated as: {eq}\text{ moles of Sn}^{2+} = \frac{\text{ moles of KMnO}_{4}}{5} {/eq} The number of moles of {eq}KMnO_{4} {/eq} used in the titration is: {eq}\text{ moles of KMnO}_{4} = \text{ concentration} times \text{ volume} = 0.100 \text{ mol/L} times 40.1 \text{ mL} = 0.00401 \text{ mol} {/eq} Substituting this into the equation above, we get: {eq}\text{ moles of Sn}^{2+} = \frac{0.00401 \text{ mol}}{5} = 0.000802 \text{ mol} {/eq} The volume of the {eq}Sn^{2+} {/eq} solution used in the titration is 25.00 mL, which is equivalent to 0.02500 L. Therefore, the concentration of the {eq}Sn^{2+} {/eq} solution can be calculated as: {eq}\text{ concentration} = \frac{\text{ moles}}{\text{ volume}} = \frac{0.000802 \text{ mol}}{0.02500 \text{ L}} = boxed{0.03208 \text{ mol/L}} {/eq} Therefore, the concentration of the {eq}Sn^{2+}(aq) {/eq} in mol/L is 0.03208.