Problem 73 - Chemistry and Chemical Reactivity

Question

If you mix \(48 \mathrm{mL}\) of \(0.0012 \mathrm{M} \mathrm{BaCl}_{2}\) with \(24 \mathrm{mL}\) of \(1.0 \times 10^{-6} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4},\) will a precipitate of \(\mathrm{BaSO}_{4}\) form?

Step-by-Step Solution

Answer

A precipitate of BaSO\(_4\) will form because \(Q_{sp} > K_{sp}\).

Step 1: Calculate moles of each reactant

We first need to calculate the moles of each reactant. For BaCl\(_2\), the number of moles is calculated using its concentration and volume: \(\text{moles of BaCl}_2 = 0.0012 \, M \times 0.048 \, L = 5.76 \times 10^{-5} \, \text{moles}\). For Na\(_2\)SO\(_4\), the moles are \(\text{moles of Na}_2\text{SO}_4 = 1.0 \times 10^{-6} \, M \times 0.024 \, L = 2.4 \times 10^{-8} \, \text{moles}\).

Step 2: Determine the limiting reactant

The balanced chemical equation for the reaction is \(\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl}\). The stoichiometry shows a 1:1 ratio between BaCl\(_2\) and Na\(_2\)SO\(_4\). Since \(5.76 \times 10^{-5} \, \text{moles} > 2.4 \times 10^{-8} \, \text{moles}\), Na\(_2\)SO\(_4\) is the limiting reactant.

Step 3: Calculate the concentration of Ba\(^{2+}\) and SO\(_4^{2-}\) in the mixture

To find the concentrations, we calculate the total volume of the mixture, which is \(48 \, \text{mL} + 24 \, \text{mL} = 72 \, \text{mL} = 0.072 \, \text{L}\). The concentration of Ba\(^{2+}\) ions is \(\frac{5.76 \times 10^{-5}}{0.072} = 8 \times 10^{-4} \, M\). The concentration of SO\(_4^{2-}\) ions is \(\frac{2.4 \times 10^{-8}}{0.072} = 3.33 \times 10^{-7} \, M\).

Step 4: Calculate the reaction quotient \(Q_{sp}\)

The reaction quotient \(Q_{sp}\) is given by \(Q_{sp} = [\text{Ba}^{2+}] \times [\text{SO}_4^{2-}] = (8 \times 10^{-4} \, M) \times (3.33 \times 10^{-7} \, M) = 2.664 \times 10^{-10}\).

Step 5: Compare \(Q_{sp}\) with \(K_{sp}\) of BaSO\(_4\)

The solubility product constant \(K_{sp}\) for BaSO\(_4\) is approximately \(1.1 \times 10^{-10}\). Since \(Q_{sp} = 2.664 \times 10^{-10} > K_{sp} = 1.1 \times 10^{-10}\), a precipitate will form.

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that gets completely consumed first. This reactant limits the amount of product that can be formed. To identify the limiting reactant, we look at the initial amount of each reactant and the stoichiometry from the balanced chemical equation.

In the given exercise, we mixed barium chloride (BaCl _2 ) and sodium sulfate (Na _2 SO _4 ) to form barium sulfate (BaSO _4 ) and sodium chloride (NaCl). The equation is balanced with a 1:1 molar ratio between BaCl _2 and Na _2 SO _4 .

After calculating, sodium sulfate was found to have fewer moles (2.4 x 10⁻⁸ moles) compared to barium chloride (5.76 x 10⁻⁵ moles). Hence, sodium sulfate is the limiting reactant. It determines the maximum amount of BaSO _4 that can form since it will be used up first in the reaction.

Reaction Quotient

The reaction quotient, denoted as Q, is a measure used to understand the direction in which a reaction will proceed. It compares the present concentrations of products and reactants to the reaction's equilibrium conditions.

For precipitation reactions, we use the reaction quotient for solubility, often written as Q_sp. It's calculated using the ion concentrations involved in forming a precipitate. By contrasting Q_sp with the solubility product constant (K_sp), we can determine if a precipitate will form.

In our scenario, we calculate Q_sp by multiplying the concentrations of Ba ^{2+} and SO _4^{2-} ions in the solution. If Q_sp is greater than K_sp, it indicates that the product concentrations exceed their equilibrium values, and a precipitate will form. Conversely, if Q_sp is less than K_sp, no precipitate forms, as the solution remains unsaturated.

Solubility Product Constant

The solubility product constant, abbreviated as K_sp, is a specific type of equilibrium constant for a saturated solution of an ionic compound. It reflects the maximum extent to which the compound can dissolve under equilibrium conditions.

K_sp is fascinated by solutions that may form a precipitate, as it defines the threshold concentration of ions in a solution. For our exercise with BaSO _4 , K_sp = 1.1 x 10⁻¹⁰ at a standard temperature. It signifies the product of the molar concentration of the constituent ions when in equilibrium.

When the reaction quotient Q_sp exceeds the K_sp value, the ions exceed their solubility limit, resulting in precipitate formation. This vital concept helps predict whether an ionic compound will remain dissolved or start to settle out of the solution.

Ion Concentration Calculation

Understanding how to calculate the concentration of ions in a solution is key to exploring many chemical phenomena, including precipitation reactions. Concentration is the amount of a substance in a given volume of solution, typically expressed in molarity (M).

To find ion concentrations, one can use the formula: \[ ext{Molarity (M)} = rac{ ext{Moles of solute}}{ ext{Liters of solution}} \]

In the exercise, we mixed solutions of BaCl_2 and Na_2SO_4. After mixing, the total solution volume was 0.072 L. For barium ions, we calculated the concentration as:\[ ext{[Ba}^{2+} ext{]} = rac{5.76 imes 10^{-5}}{0.072} = 8 imes 10^{-4} ext{M} \] Similarly, for sulfate ions:\[ ext{[SO}_4^{2-} ext{]} = rac{2.4 imes 10^{-8}}{0.072} = 3.33 imes 10^{-7} ext{M} \] These calculations allowed us to determine Q_sp and conclude about the potential formation of a BaSO_4 precipitate.