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Under basic conditions, \(\mathrm{MnO}_{4}^{-}\) is used as a titrant for the analysis of \(\mathrm{Mn}^{2+}\), with both the analyte and the titrant forming \(\mathrm{MnO}_{2}\). In the analysis of a mineral sample for manganese, a \(0.5165-\mathrm{g}\) sample is dissolved and the manganese reduced to \(\mathrm{Mn}^{2+}\). The solution is made basic and titrated with \(0.03358 \mathrm{M} \mathrm{KMnO}_{4}\), requiring \(34.88 \mathrm{~mL}\) to reach the end point. Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{Mn}\) in the mineral sample.

Short Answer

Expert verified
The \\(m{ ext{percentage by weight}}\\) of manganese in the sample is 18.67%.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between permanganate ion \(\mathrm{MnO}_{4}^{-}\) and manganese (II) ion \(\mathrm{Mn}^{2+}\) under basic conditions is: \[2\mathrm{MnO}_{4}^{-} + 3\mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \rightarrow 5\mathrm{MnO}_{2} + 4\mathrm{OH}^{-}\]This equation shows that 2 moles of \(\mathrm{MnO}_{4}^{-}\) react with 3 moles of \(\mathrm{Mn}^{2+}\) to form 5 moles of \(\mathrm{MnO}_{2}\).
02

Calculate Moles of \(\mathrm{KMnO}_{4}\) Used

Use the concentration and volume of \(\mathrm{KMnO}_{4}\) solution to find the moles of permanganate ion used:\[moles \ of \ \mathrm{KMnO}_{4} = 0.03358 \ \mathrm{M} \times \frac{34.88 \ \mathrm{mL}}{1000 \ \mathrm{mL/L}} = 0.001169 \ \mathrm{mol}\]
03

Relate Moles of \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\)

From the balanced equation, 2 moles of \(\mathrm{MnO}_{4}^{-}\) react with 3 moles of \(\mathrm{Mn}^{2+}\). Therefore, the moles of \(\mathrm{Mn}^{2+}\) reacted are:\[moles \ of \ \mathrm{Mn}^{2+} = \frac{3}{2} \times 0.001169 \ \mathrm{mol} = 0.0017535 \ \mathrm{mol}\]
04

Calculate Mass of \(\mathrm{Mn}^{2+}\)

The molar mass of manganese \(\mathrm{Mn}\) is approximately \(54.94 \ \mathrm{g/mol}\). Thus, the mass of \(\mathrm{Mn}^{2+}\) can be calculated as:\[mass \ of \ \mathrm{Mn} = 0.0017535 \ \mathrm{mol} \times 54.94 \ \mathrm{g/mol} = 0.09639 \ \mathrm{g}\]
05

Calculate \\(m{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{ ext{

The percent by weight of manganese \(\mathrm{Mn}\) in the mineral sample is:\[\% \mathrm{w}/\mathrm{w} \ \mathrm{Mn} = \left(\frac{0.09639 \ \mathrm{g}}{0.5165 \ \mathrm{g}}\right) \times 100 \% = 18.67 \%\]
06

Conclusion

Thus, the percentage by weight of manganese in the mineral sample is 18.67%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permanganate titration
Permanganate titration is a particularly useful method in chemistry for analyzing manganese compounds. Permanganate ions (\(\mathrm{MnO}_{4}^{-}\)) serve as titrants, transforming in a visually evident reaction. The purple color of permanganate diminishes as it reacts, indicating the endpoint.

This process is often performed under basic conditions to ensure complete reaction and stability. When you perform a titration, you add a titrant (a solution of known concentration) to the analyte (the substance you're analyzing) until the reaction is complete.

Using permanganate ions, the analysis of manganese (\(\mathrm{Mn}^{2+}\)) allows for the accurate determination of its concentration based on the changes in color. In this reaction, each stage is precise and requires careful observation to ensure accurate results.
Percent composition calculation
Percent composition calculation helps determine how much of a specific element is present in a compound as a percentage. It's essential in analyzing samples such as minerals to determine their composition.

To calculate it, you first need to know the mass of the element in question and the total mass of the sample. The formula is simple:
  • \[\% \text{w/w} = \left( \frac{\text{mass of element}}{\text{total mass of sample}} \right) \times 100 \%\]
After determining the mass of manganese (\(0.09639 \ \mathrm{g}\)) from the titration, and knowing the initial sample mass (\(0.5165 \ \mathrm{g}\)), you substitute these values into the formula. This gives you a manganese percentage weight of \(18.67\%\). Calculating percent composition is critical for quality control and verifying the properties of mineral samples with specific industrial and economic uses.
Balanced chemical equation
A balanced chemical equation ensures the conservation of mass by having equal numbers of each atom on both sides of the reaction. Balancing is crucial for accurately calculating reactants and products in any chemical reaction.

In this exercise, the balanced equation:\[2 \mathrm{MnO}_{4}^{-} + 3 \mathrm{Mn}^{2+} + 2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{MnO}_{2} + 4 \mathrm{OH}^{-}\]shows the stoichiometry of the reaction.
  • It tells us that every 2 moles of permanganate ion reacts with 3 moles of manganese ion to produce 5 moles of manganese dioxide.
  • The balanced equation also highlights the involvement of water and hydroxide ions, reflecting the basic conditions needed for the process.
Understanding how to balance equations is fundamental for solving stoichiometric problems and predicting the outcomes of reactions.

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Most popular questions from this chapter

The concentration of \(o\) -phthalic acid in an organic solvent, such as \(n\) butanol, is determined by an acid-base titration using aqueous \(\mathrm{NaOH}\) as the titrant. As the titrant is added, the \(o\) -phthalic acid extracts into the aqueous solution where it reacts with the titrant. The titrant is added slowly to allow sufficient time for the extraction to take place. (a) What type of error do you expect if the titration is carried out too quickly? (b) Propose an alternative acid-base titrimetric method that allows for a more rapid determination of the concentration of \(o\) -phthalic acid in \(n\) -butanol.

The concentration of \(\mathrm{SO}_{2}\) in air is determined by bubbling a sample of air through a trap that contains \(\mathrm{H}_{2} \mathrm{O}_{2} .\) Oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\) results in the formation of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is then determined by titrating with \(\mathrm{NaOH}\). In a typical analysis, a sample of air is passed through the peroxide trap at a rate of \(12.5 \mathrm{~L} / \mathrm{min}\) for \(60 \mathrm{~min}\) and required \(10.08 \mathrm{~mL}\) of \(0.0244 \mathrm{M} \mathrm{NaOH}\) to reach the phenolphthalein end point. Calculate the \(\mu \mathrm{L} / \mathrm{L} \mathrm{SO}_{2}\) in the sample of air. The density of \(\mathrm{SO}_{2}\) at the temperature of the air sample is \(2.86 \mathrm{mg} / \mathrm{mL}\).

Explain why it is not possible for a sample of water to simultaneously have \(\mathrm{OH}^{-}\) and \(\mathrm{HCO}_{3}^{-}\) as sources of alkalinity.

Calculate or sketch titration curves for the following redox titration reactions at \(25^{\circ} \mathrm{C}\). Assume the analyte initially is present at a concentration of \(0.0100 \mathrm{M}\) and that a \(25.0-\mathrm{mL}\) sample is taken for analysis. The titrant, which is the underlined species in each reaction, has a concentration of \(0.0100 \mathrm{M}\). (a) \(\mathrm{V}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{V}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)\) (b) \(\mathrm{Sn}^{2+}(a q)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)+2 \mathrm{Ce}^{3+}(a q)\) (c) \(5 \mathrm{Fe}^{2+}(a q)+\underline{\mathrm{MnO}_{4}^{-}(a q)}+8 \mathrm{H}^{+}(a q) \longrightarrow\) $$ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(\iota)(\text { at } \mathrm{pH}=1) $$

Schwartz published the following simulated data for the titration of a \(1.02 \times 10^{-4} \mathrm{M}\) solution of a monoprotic weak acid \(\left(\mathrm{p} K_{\mathrm{a}}=8.16\right)\) with \(1.004 \times 10^{-3} \mathrm{M} \mathrm{NaOH} .{ }^{10}\) The simulation assumes that a \(50-\mathrm{mL}\) pipet is used to transfer a portion of the weak acid solution to the titration vessel. A calibration of the pipet shows that it delivers a volume of only 49.94 mL. Prepare normal, first derivative, second derivative, and Gran plot titration curves for this data, and determine the equivalence point for each. How do these equivalence points compare to the expected equivalence point? Comment on the utility of each titration curve for the analysis of very dilute solutions of very weak acids. $$ \begin{array}{cccc} \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} & \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} \\ \hline 0.03 & 6.212 & 4.79 & 8.858 \\ 0.09 & 6.504 & 4.99 & 8.926 \end{array} $$ $$ \begin{array}{cccc} \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} & \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} \\ \hline 0.29 & 6.936 & 5.21 & 8.994 \\ 0.72 & 7.367 & 5.41 & 9.056 \\ 1.06 & 7.567 & 5.61 & 9.118 \\ 1.32 & 7.685 & 5.85 & 9.180 \\ 1.53 & 7.776 & 6.05 & 9.231 \\ 1.76 & 7.863 & 6.28 & 9.283 \\ 1.97 & 7.938 & 6.47 & 9.327 \\ 2.18 & 8.009 & 6.71 & 9.374 \\ 2.38 & 8.077 & 6.92 & 9.414 \\ 2.60 & 8.146 & 7.15 & 9.451 \\ 2.79 & 8.208 & 7.36 & 9.484 \\ 3.01 & 8.273 & 7.56 & 9.514 \\ 3.19 & 8.332 & 7.79 & 9.545 \\ 3.41 & 8.398 & 7.99 & 9.572 \\ 3.60 & 8.458 & 8.21 & 9.599 \\ 3.80 & 8.521 & 8.44 & 9.624 \\ 3.99 & 8.584 & 8.64 & 9.645 \\ 4.18 & 8.650 & 8.84 & 9.666 \\ 4.40 & 8.720 & 9.07 & 9.688 \\ 4.57 & 8.784 & 9.27 & 9.706 \end{array} $$

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