Question

Consider a metal ion \(\mathrm{A}^{2+}\) and its nitrate salt, \(\mathrm{A}\left(\mathrm{NO}_{3}\right)_{2}\). In an experiment, \(35.00 \mathrm{~mL}\) of a \(0.217 \mathrm{M}\) solution of \(\mathrm{A}\left(\mathrm{NO}_{3}\right)_{2}\) is made to react with \(25.00 \mathrm{~mL}\) of \(0.195 \mathrm{M} \mathrm{NaOH} .\) A precipitate, \(\mathrm{A}(\mathrm{OH})_{2},\) forms. Along with the precipitation, the temperature increases from \(24.8^{\circ} \mathrm{C}\) to \(28.2^{\circ} \mathrm{C}\). What is \(\Delta H\) for the precipitation of \(\mathrm{A}(\mathrm{OH})_{2}\) ? The following assumptions can be made. The density of the solution is \(1.00 \mathrm{~g} / \mathrm{mL}\). Volumes are additive. The specific heat of the solution is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).

Short answer

Answer: The enthalpy change for the precipitation of A(OH)₂ is 112.2 kJ/mol.

Step-by-step solution

Calculate the moles of A(NO₃)₂ and NaOH

We can find the moles of A(NO₃)₂ and NaOH by using the given concentrations and volumes of each solution: Moles of A(NO₃)₂ = volume x concentration Moles of A(NO₃)₂ = 0.035 L x 0.217 mol/L = 0.007595 mol Moles of NaOH = volume x concentration Moles of NaOH = 0.025 L x 0.195 mol/L = 0.004875 mol

Determine the limiting reactant

The reaction will occur in a 1:2 ratio between A(NO₃)₂ and NaOH (as the metal ion A²⁺ requires two OH⁻ ions to form A(OH)₂). To find the limiting reactant, compare the moles of A(NO₃)₂ to twice the moles of NaOH: 0.007595 mol A(NO₃)₂ < 2 x 0.004875 mol NaOH (0.00975 mol) Since the available moles of A(NO₃)₂ are less than the required moles of NaOH (based on the 1:2 ratio), A(NO₃)₂ is the limiting reactant.

Calculate the heat (q) released in the reaction

We can use the specific heat, mass, and temperature change of the solution to find the heat released during the reaction: q = mass x specific heat x ΔT q = (35.00 mL + 25.00 mL) x 1.00 g/mL x 4.18 J/(g·°C) x (28.2°C - 24.8°C) q = 60.00 g x 4.18 J/(g·°C) x 3.4°C = 852.72 J

Calculate ΔH for the precipitation of A(OH)₂

To find ΔH, divide the heat (q) released by the moles of A(OH)₂ that formed during the reaction. Using the moles of the limiting reactant A(NO₃)₂ (0.007595 mol), we can calculate the moles of A(OH)₂ formed: ΔH = q / moles of A(OH)₂ ΔH = 852.72 J / 0.007595 mol = 112217 J/mol Therefore, the enthalpy change for the precipitation of A(OH)₂ is 112217 J/mol or 112.2 kJ/mol.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is fundamental to mastering chemistry. Stoichiometry is the section of chemistry that pertains to the quantitative relationships between the substances as they participate in chemical reactions. When a chemical reaction occurs, reactants are transformed into products with definite proportions that follow the law of conservation of mass. For a balanced chemical reaction, the mole ratio of reactants and products can be calculated, and this ratio is essential for solving stoichiometric problems.

In the given exercise, stoichiometry comes into play when calculating the moles of the metal nitrate salt, \(A(NO_3)_2\), and the sodium hydroxide, \(NaOH\), using their respective molarities and volumes. The balanced chemical equation gives us the stoichiometric coefficients that define the proportions in which reactants combine. This insight is crucial for the next steps, especially in determining the limiting reactant, which then informs us about the quantity of product formed in the reaction.

Limiting Reactant Determination

The concept of limiting reactant is critical in stoichiometry and real-world applications, such as manufacturing and laboratory experiments. The limiting reactant in a chemical reaction is the substance that is completely consumed first and thus limits the amount of products formed. Identifying the limiting reactant is important because it dictates the maximum yield of the reaction.

To determine the limiting reactant, one must first understand the mole ratio of the reactants involved, as provided by the balanced equation. In our exercise, the metal ion \(A^{2+}\) reacts with hydroxide ions \(OH^-\) in a 1:2 ratio to form a precipitate of \(A(OH)_2\). By comparing the moles we calculated for both \(A(NO_3)_2\) and \(NaOH\) and taking into account this mole ratio, we discover that \(A(NO_3)_2\) is the limiting reactant. This understanding enables us to calculate the theoretical yield of \(A(OH)_2\) precipitate formed and the heat released during the process.

Heat Released in Reactions

The heat released or absorbed during a chemical reaction is an important aspect of thermodynamics and is known as enthalpy change, denoted as \(\Delta H\). The enthalpy change can be either positive or negative, indicating whether a reaction is endothermic (absorbs heat) or exothermic (releases heat). In practical terms, quantifying this heat change involves measurement of the temperature change of the system, assuming constant pressure, along with knowledge of the system's mass and specific heat capacity.

In the precipitation reaction of our example, the temperature increase observed is a sign that it's an exothermic process. By using the formula \(q = mass \times specific heat \times \Delta T\), we can calculate the total heat \(q\) released into the surroundings. Once calculated, we then normalize this heat quantity to one mole of product to determine the molar enthalpy change, \(\Delta H\), for the formation of the precipitate \(A(OH)_2\). This value helps us understand the energy dynamics of the precipitation reaction and is directly related to the strength of the bonds formed in the product campared to the reactant bonds broken.