Problem 55 - Chemistry: The Central Science in SI Units, Global Edition

Question

A 1.50-L solution saturated at \(25^{\circ} \mathrm{C}\) with cobalt carbonate \(\left(\mathrm{CoCO}_{3}\right)\) contains \(2.71 \mathrm{mg}\) of \(\mathrm{CoCO}_{3} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The solubility product constant for cobalt carbonate (CoCO3) at 25°C, calculated from a 1.50 L solution saturated with 2.71 mg of CoCO3, is \( K_{sp} = 2.314 \times 10^{-10} \).

Step 1: Convert the mass of CoCO3 to moles

We are given that the solution contains 2.71 mg of CoCO3. We need to convert this mass into moles. To do this, we'll use the molar mass of CoCO3, which can be calculated by adding the molar masses of its individual components: Co (58.93 g/mol), C (12.01 g/mol), and O3 (3 × 16.00 g/mol). The molar mass of CoCO3 is: CoCO3 = 58.93 + 12.01 + (3 × 16.00) = 118.94 g/mol Now we can convert the mass of CoCO3 into moles: moles of CoCO3 = mass of CoCO3 / molar mass of CoCO3 moles of CoCO3 = 2.71 mg × (1 g / 1000 mg) × (1 mol / 118.94 g) moles of CoCO3 = 2.282 × 10^{-5} mol

Step 2: Calculate the molar concentration of CoCO3

We are given that the volume of the solution is 1.50 L. To find the molar concentration of CoCO3, we will divide the moles of CoCO3 by the volume of the solution in liters: [CoCO3] = moles of CoCO3 / volume of solution [CoCO3] = 2.282 × 10^{-5} mol / 1.50 L [CoCO3] = 1.522 × 10^{-5} M

Step 3: Calculate the solubility product constant, Ksp

Now we can use the molar concentration of CoCO3 to calculate its solubility product constant, Ksp. The solubility product constant is given by: Ksp = [Co²⁺][CO3²⁻] Since one mole of CoCO3 dissociates into one mole of Co²⁺ and one mole of CO3²⁻, their molar concentrations will be equal. [Co²⁺] = [CO3²⁻] = 1.522 × 10^{-5} M Now we can calculate Ksp: Ksp = (1.522 × 10^{-5})(1.522 × 10^{-5}) Ksp = 2.314 × 10^{-10} The solubility product constant for cobalt carbonate (CoCO3) at 25°C is 2.314 × 10^{-10}.

Key Concepts

Cobalt Carbonate

Cobalt Carbonate, known chemically as \( \text{CoCO}_3 \), is a compound comprised of cobalt, carbon, and oxygen. This inorganic salt appears as a pink powder and is often used in industrial applications such as ceramics and glass manufacturing. At a molecular level, cobalt carbonate consists of one cobalt (Co) atom, one carbon (C) atom, and three oxygen (O) atoms. It is not typically found in water but can dissolve under certain conditions, forming a saturated solution. This behavior makes it an interesting subject for investigations involving solubility equilibria.

Molar Mass Calculation

Calculating the molar mass of a compound is essential for converting mass to moles, which is a crucial step in solving many chemical problems. The molar mass of cobalt carbonate \( \text{CoCO}_3 \) can be found by summing up the atomic masses of its constituent elements, carefully adding together the mass of cobalt (\(58.93 \text{ g/mol}\)), carbon (\(12.01 \text{ g/mol}\)), and oxygen. Since there are three oxygens in the molecular formula, we calculate:
\[ \text{Molar mass of CoCO}_3 = 58.93 + 12.01 + (3 \times 16.00) = 118.94 \text{ g/mol} \]
This calculated molar mass is crucial as it allows us to convert the tiny mass of the compound (such as 2.71 mg) into moles, aiding in determining its concentration in solution.

Molar Concentration

Molar concentration, often represented as molarity (M), measures the number of moles of a solute within a given volume of solution. It is calculated by dividing the moles of solute by the volume of the solution in liters. In the case of cobalt carbonate:
\[ \text{Moles of } \text{CoCO}_3 = \frac{\text{Mass of } \text{CoCO}_3}{\text{Molar mass of } \text{CoCO}_3} \] \[ \text{Moles of } \text{CoCO}_3 = \frac{2.71 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}}}{118.94 \text{ g/mol}} = 2.282 \times 10^{-5} \text{ mol} \]
Given that the solution's volume is 1.50 L, the molar concentration is:
\[ [\text{CoCO}_3] = \frac{2.282 \times 10^{-5} \text{ mol}}{1.50 \text{ L}} = 1.522 \times 10^{-5} \text{ M} \]
Knowing this concentration is vital, as it lays the groundwork for subsequent calculations involving chemical equilibrium and solubility.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. In the context of solubility, we're interested in the equilibrium established when a compound, such as cobalt carbonate, dissolves in a solution. For \( \text{CoCO}_3 \), when it dissolves, it breaks down into its ions: cobalt \( \text{Co}^{2+} \) and carbonate \( \text{CO}_3^{2-} \). This dissociation is key in determining the solubility product constant \( K_{sp} \).
The equilibrium expression for cobalt carbonate in a saturated solution is:
\[ K_{sp} = [\text{Co}^{2+}][\text{CO}_3^{2-}] \]
Since one mole of \( \text{CoCO}_3 \) produces one mole of \( \text{Co}^{2+} \) and one mole of \( \text{CO}_3^{2-} \), the concentrations are equal. In equilibrium, these values give the \( K_{sp} \) as:
\[ K_{sp} = (1.522 \times 10^{-5})^2 = 2.314 \times 10^{-10} \]
Understanding \( K_{sp} \) helps predict whether, under certain conditions, more \( \text{CoCO}_3 \) will dissolve or precipitate, also informing us about the solubility limits of the compound in water.