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Chapter 6: Problem 14

A particular acid-rain water has \(\mathrm{SO}_{3}{ }^{2-}\). If a \(25.00 \mathrm{~mL}\) sample of this water requires \(34.08 \mathrm{~mL}\) of \(0.01964 \mathrm{M} \mathrm{KMnO}_{4}\) for its titration, what is the molarity of \(\mathrm{SO}_{3}{ }^{2-}\) in acid-rain? \(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{SO}_{3}^{2-}+6 \mathrm{H}^{+} \longrightarrow 5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The molarity of \text{SO}_{3}{ }^{2-} in the acid-rain sample is approximately 0.0669 M.

Step by step solution

01

Identify the reaction and balance it

Given the equation: 2 MnO4- + 5 SO3^{2-} + 6 H+ -> 5 SO4^{2-} + 2 Mn^{2+} + 3 H2O. Ensure that the reaction is balanced; in this case, it is already balanced as provided.
02

Calculate moles of KMnO4 used in the titration

Use the molarity and volume of KMnO4 to calculate moles. Molarity (M) = moles/L, thus moles = Molarity * Volume. Moles of KMnO4 = 0.01964 M * 34.08 mL * (1 L / 1000 mL) = 6.692192e-4 moles.
03

Determine the moles of SO3^{2-} reacted

Using stoichiometry from the balanced equation, calculate moles of SO3^{2-}. From the reaction, 2 moles of KMnO4 react with 5 moles of SO3^{2-}. Therefore, moles of SO3^{2-} = (5 moles SO3^{2-} / 2 moles KMnO4) * 6.692192e-4 moles KMnO4 = 1.673048e-3 moles SO3^{2-}.
04

Calculate the molarity of SO3^{2-} in the sample

Molarity (M) of SO3^{2-} = moles of SO3^{2-} / Volume of sample in liters. Volume of the sample is 25.00 mL which is 0.025 L. Molarity = 1.673048e-3 moles / 0.025 L = 0.06692192 M.

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Key Concepts

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

Titration Calculation

Understanding titration calculation is essential for many chemistry applications, including determining concentrations of substances in solutions. In the context of our acid-rain titration problem, the key objective is to use the titration process to find the unknown concentration of sulfate ions (\text{SO}_3^{2-}) in an acid-rain sample. To do this, a standard solution of potassium permanganate (\text{KMnO}_4) with a known molarity is gradually added to the acid-rain sample until the reaction between \text{KMnO}_4 and \text{SO}_3^{2-} is complete, a point known as the endpoint.

Volume and Molarity Relationship

The volume of the titrant (\text{KMnO}_4) used provides information on how many moles of \text{SO}_3^{2-} were present in the sample. In this specific problem, the volume of \text{KMnO}_4 used in the titration is \text{34.08 mL}, and its concentration is \text{0.01964 M}. To calculate the moles of \text{KMnO}_4, apply the formula:

  • \text{Moles} = \text{Molarity} \( \times \) \text{Volume}

By using the titration calculation, we establish how these volumes and molarity provide a pathway to finding the concentration of the unknown substance.

Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. In stoichiometric calculations, the coefficients from a balanced equation are used to set up ratios, enabling us to convert between moles of different substances involved in the reaction.

Stoichiometric Ratios in Titration

In our exercise, the balanced equation shows that 2 moles of \text{KMnO}_4 react with 5 moles of \text{SO}_3^{2-}. With stoichiometry, we use this ratio to determine how many moles of \text{SO}_3^{2-} react based on the moles of \text{KMnO}_4. The calculation is:

  • \text{Moles of} \text{SO}_3^{2-} = (\(\frac{5 \text{ moles} \text{SO}_3^{2-}}{2 \text{ moles} \text{KMnO}_4}\)) \(\times\) \text{Moles of} \text{KMnO}_4

This ratio is the heart of the stoichiometry involved in our titration problem. Understanding this ratio is crucial for determining the exact amount of \text{SO}_3^{2-} that has participated in the chemical reaction, and thus the molarity of \text{SO}_3^{2-} in the sample.

Molarity

Molarity, often represented by the capital M, is a measure of concentration that describes the number of moles of a solute per liter of solution. It's a fundamental concept in chemistry, allowing scientists to predict how substances will react with one another in a solution. In the case of acid-rain analysis, the molarity of \text{SO}_3^{2-} will indicate the level of pollution in the water sample.

Calculating Molarity from Titration Data

To calculate the molarity of \text{SO}_3^{2-} using our titration data, we take the moles of \text{SO}_3^{2-} found through stoichiometry and divide by the volume of the acid-rain water sample in liters. In mathematical terms:

  • \text{Molarity of} \text{SO}_3^{2-} = \(\frac{\text{Moles of} \text{SO}_3^{2-}}{\text{Volume of sample in liters}}\)

After the step by step calculations have been performed, we arrive at the molarity of the \text{SO}_3^{2-} in the acid-rain sample, concluding our titration problem.

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Most popular questions from this chapter

\(25 \mathrm{~mL}\) of \(0.017 \mathrm{H}_{2} \mathrm{SO}_{3}\) in strongly acidic medium required \(16.9 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{KMnO}_{4}\) and in neutral medium required \(28.6 \mathrm{~mL}\) of \(0.01 M\) \(\mathrm{KMnO}_{4}\) for complete conversion of \(\mathrm{SO}_{3}{ }^{2-}\) to \(\mathrm{SO}_{4}^{2-}\). Assign the oxidation no. of Mn in the product formed in each case.

A \(1.100 \mathrm{~g}\) sample of copper ore is dissolved and the \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is treated with excess \(\mathrm{KI}\). The liberated \(\mathrm{I}_{2}\) requires \(12.12 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution for titration. What is \(\%\) copper by mass in the ore?

\(\mathrm{Mn}^{2+}(\) aq \()\) can be determined by titration with \(\mathrm{MnO}_{4}^{-}(\) aq \()\) $$ 3 \mathrm{Mn}^{2+}+2 \mathrm{MnO}_{4}^{-} \longrightarrow 6 \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ A \(25.00 \mathrm{~mL}\) sample of \(\mathrm{Mn}^{2+}\) (aq) requires \(34.77 \mathrm{~mL}\) of \(0.05876 M\) \(\mathrm{KMnO}_{4}\) (aq) for its titration. What is the molarity of the \(\mathrm{Mn}^{2+}\) (aq)?

Reaction, \(2 \mathrm{Br}^{-}{ }_{(\mathrm{aq})}+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}_{(\mathrm{aq} .)}+\mathrm{Br}_{2(\mathrm{aq} .)}\), is used for commercial preparation of bromine from its salts. Suppose we have \(50 \mathrm{~mL}\) of a \(0.060 \mathrm{M}\) solution of NaBr. What volume of a \(0.050 \mathrm{M}\) solution of \(\mathrm{Cl}_{2}\) is needed to react completely with the Br?
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