Problem 110 - Chemistry and Chemical Reactivity

Question

A The amount of oxygen, \(\mathrm{O}_{2}\), dissolved in a water sample at \(25^{\circ} \mathrm{C}\) can be determined by titration. The first step is to add solutions of \(\mathrm{MnSO}_{4}\) and NaOH to the water to convert the dissolved oxygen to \(\mathrm{MnO}_{2}\). A solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{KI}\) is then added to convert the \(\mathrm{MnO}_{2}\) to \(\mathrm{Mn}^{2+},\) and the iodide ion is converted to \(\mathrm{I}_{2}\). The \(\mathrm{I}_{2}\) is then titrated with standardized \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3-}\) (a) Balance the equation for the reaction of \(\mathrm{Mn}^{2+}\) ions with \(\mathrm{O}_{2}\) in basic solution. (b) Balance the equation for the reaction of \(\mathrm{MnO}_{2}\) with \(\mathbf{I}^{-}\) in acid solution. (c) Balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{2}\) (d) Calculate the amount of \(\mathbf{O}_{2}\) in \(25.0 \mathrm{mL}\) of water if the titration requires \(2.45 \mathrm{mL}\) of \(0.0112 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution.

Step-by-Step Solution

Answer

The mass of \(\text{O}_2\) in the sample is \(4.390 \times 10^{-4}\) g.

Step 1: Balance Mn2+ with O2 in Basic Solution

In basic solution, the reaction of \(\text{Mn}^{2+}\) with \(\text{O}_{2}\) forms \(\text{MnO}_{2}\). The unbalanced equation is: \[\text{Mn}^{2+} (aq) + \text{O}_{2} (g) \rightarrow \text{MnO}_{2} (s)\]Balancing in basic solution:- Write half-reactions and balance atoms other than \(\text{O}\) and \(\text{H}\).- Balance oxygen by adding water:\[4\text{OH}^- (aq) + \text{Mn}^{2+} (aq) + \text{O}_{2} (g) \rightarrow \text{MnO}_{2} (s) + 2\text{H}_2\text{O} (l)\]- Balance charge by adding \(\text{OH}^-\):\[2\text{Mn}^{2+} (aq) + \text{O}_{2} (g) + 4\text{OH}^- (aq) \rightarrow 2\text{MnO}_{2} (s)+2\text{H}_2\text{O} (l)\]This is the balanced equation.

Step 2: Balance MnO2 with I- in Acidic Solution

The conversion of \(\text{MnO}_{2}\) to \(\text{Mn}^{2+}\) in acidic solution with \(\text{I}^-\) also forms \(\text{I}_{2}\). The unbalanced equation is:\[\text{MnO}_{2} (s) + \text{I}^- (aq) \rightarrow \text{Mn}^{2+} (aq) + \text{I}_{2} (aq)\]Balancing in acidic solution:- Write half-reactions:\[\text{MnO}_{2} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} \]\[2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-\]- Multiply to equalize electrons and add:\[\text{MnO}_{2} + 4\text{H}^+ + 2\text{I}^- \rightarrow \text{Mn}^{2+} + \text{I}_2 + 2\text{H}_2\text{O}\]This is the balanced equation.

Step 3: Balance S2O32- with I2

In this reaction, \(\text{S}_{2}\text{O}_{3}^{2-}\) reacts with \(\text{I}_{2}\) to form \(\text{I}^-\):\[\text{I}_{2} + \text{S}_{2}\text{O}_{3}^{2-} \rightarrow \text{I}^- + \text{S}_{4}\text{O}_{6}^{2-}\]Balancing the charges:- Balance each side by multiplying the iodide reaction by 2:\[\text{I}_{2} + 2\text{S}_{2}\text{O}_{3}^{2-} \rightarrow 2\text{I}^- + \text{S}_{4}\text{O}_{6}^{2-}\]This is the balanced equation.

Step 4: Calculate Moles of Na2S2O3

Using the volume and molarity:\[\text{Moles of Na}_2\text{S}_2\text{O}_3 = 2.45 \text{ mL} \times 0.0112 \text{ M}\]Convert mL to L:\[0.00245 \text{ L} \times 0.0112 \text{ M} = 2.744 \times 10^{-5} \text{ moles}\]

Step 5: Find Moles of I2 and Corresponding O2

From the balanced equation, 1 mole of \(\text{I}_{2} \) reacts with 2 moles of \(\text{S}_{2}\text{O}_{3}^{2-} \). Therefore, the moles of \(\text{I}_{2}\) is half the moles of \(\text{S}_{2}\text{O}_{3}^{2-} \):\[\text{Moles of I}_2 = \frac{2.744 \times 10^{-5}}{2} = 1.372 \times 10^{-5}\]From the balanced equation in Step 1, each mole of \(\text{O}_{2} \) produces 1 mole of \(\text{I}_{2} \), so moles of \(\text{O}_{2} \):\[1.372 \times 10^{-5} \text{ moles}\]

Step 6: Calculate Mass of O2

The molar mass of \(\text{O}_{2}\) is 32.00 g/mol:\[\text{Mass of O}_{2} = 1.372 \times 10^{-5} \text{ moles} \times 32.00 \text{ g/mol} = 4.390 \times 10^{-4} \text{ g}\]

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances through breaking and forming of chemical bonds. They are fundamental to the study of chemistry and occur when reactants interact to form products. This can be represented as a chemical equation where the substances on the left side of the arrow (reactants) are transformed into those on the right side (products).
In the given exercise, several chemical reactions occur in different conditions - basic and acidic. Here, understanding the conditions is crucial, such as basic solutions involving hydroxide ions ( OH⁻ ) and acidic solutions involving hydrogen ions ( H⁺ ).

• In basic solutions, you often balance chemical equations by adding water molecules ( H₂O ) or hydroxide ions.
• In acidic solutions, balancing often involves adding hydrogen ions ( H⁺ ) and might require additional electrons for balance.

Recognizing these conditions helps in predicting the direction and nature of the reactions that occur.

Stoichiometry

Stoichiometry revolves around calculating the quantities of reactants and products in chemical reactions. It relies heavily on balanced chemical equations as it ensures the conservation of mass and charge. In our case, it allows us to determine how much of a product forms from a given amount of reactant.

In titration, stoichiometry is crucial in relating the titrant's volume and concentration to the amount of substance in the sample being analyzed. For instance, knowing the balanced equation for each step of the reactions in the exercise allows us to calculate precisely how much of each substance is involved.

• From the concentration of Na₂S₂O₃ and the volume used in the titration, we can calculate moles of I₂ produced.
• These calculations show that every two moles of Na₂S₂O₃ are equivalent to one mole of I₂ based on reaction stoichiometry.

Such stoichiometric calculations are essential for determining the amounts of reactants and products in a chemical reaction.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. It involves ensuring that the number of each type of atom is equal on both sides of the equation, fulfilling the conservation of mass. Balancing is achieved using coefficients before chemical formulas.

In the exercise, balancing equations under different conditions (basic and acidic) are practiced.

• For a reaction in a basic solution: First, balance metals and other elements, then balance oxygen by adding water, and complete the balancing by adding hydroxide ions to equalize charges.
• For acidic solutions, often add hydrogen ions to balance hydrogen and follow up by equalizing the charge using electrons if necessary.

Balancing ensures that the stoichiometric calculations related to the reaction will be accurate and valid. It allows us to relate moles of reactants to moles of products and is critical for quantitatively describing a chemical reaction.