Question

What is the molartiy of \(\mathrm{SO}_{4}^{2-}\) ion in aqueous solution that contain \(34.2 \mathrm{ppm}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) ? (Assume complete dissociation and density of solution \(1 \mathrm{~g} / \mathrm{mL}\) ) (a) \(3 \times 10^{-4} M\) (b) \(2 \times 10^{-4} M\) (c) \(10^{-4} M\) (d) None of these

Short answer

\(2 \times 10^{-4} M\)

Step-by-step solution

Understanding the concept of ppm

Parts per million (ppm) indicates the mass of solute (in this case, \(\mathrm{Al}_2(\mathrm{SO}_4)_3\)) divided by the total mass of the solution, multiplied by \(10^6\). We're given \(34.2\) ppm of \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\), which means there are \(34.2\) mg of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) in \(1\) liter of solution, as the density is \(1 \mathrm{g/mL}\) or \(1000 \mathrm{g/L}\).

Calculating moles of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\)

To find the molarity, we first need to determine the number of moles of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) in one liter of the solution. The molar mass of \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\) (molar mass of Al is \(27\) g/mol and S is \(32\) g/mol, O is \(16\) g/mol) is \(2 \times 27 + 3 \times (32 + 4 \times 16)\) = \(342\) g/mol. Since there are \(34.2\) mg, we convert this to grams and divide by the molar mass to get the moles.

Calculating molarity of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\)

Molarity of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) is moles of solute per liter of solution. After calculating the moles from step 2, we divide it by the volume of the solution (which is \(1\) L) to find the molarity.

Calculating molarity of \(\mathrm{SO}_4^{2-}\) ions

Each mole of \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\) dissociates into two moles of \(\mathrm{Al}^{3+}\) ions and three moles of \(\mathrm{SO}_4^{2-}\) ions. So, we need to multiply the molarity of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) by three to get the molarity of \(\mathrm{SO}_4^{2-}\) ions.

Selecting the correct answer

Find which of the given options matches the calculated molarity of \(\mathrm{SO}_4^{2-}\) in the solution.

Key concepts

Parts Per Million (ppm)

Understanding the measurement unit known as 'parts per million' (ppm) is essential when dealing with very dilute solutions in chemistry. Ppm is a way to express concentration, especially when the amounts are very small. It's a ratio of the mass of the substance (solute) to the mass of the whole solution, and it literally means 'out of a million'.

For instance, if a solution has 34.2 ppm of a substance, this means that in one million parts of this solution, 34.2 parts are the substance of interest, and the rest are the solvent or other components. To visualize this, imagine dividing a solution into one million equal parts; 34.2 of those parts would be the solute. This unit is particularly useful as it allows for easy conversion to molarity when the density of the solution is known, as it was in the given exercise with a density of 1 g/mL. Converting ppm to molarity involves calculating the mass of the solute in a given volume and then determining the number of moles of solute in that volume.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, and it is expressed in grams per mole (g/mol). The mole is a fundamental unit in chemistry that represents Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles.

To calculate molar mass, we sum the atomic masses of all the atoms in a compound's chemical formula. For a compound like \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), we add the molar masses of aluminum (Al), sulfur (S), and oxygen (O), taking into account the number of each atom in the formula. Knowing the molar mass allows chemists to convert between grams and moles, which is a key step in determination of the concentration of solutions in terms of molarity.

Dissociation of Ionic Compounds

Ionic compounds dissociate in water to form ions. This process is fundamental to understanding how these compounds behave in solution. When an ionic compound like \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) dissolves, it separates into its constituent ions - in this case, aluminum ions (\(\mathrm{Al}^{3+}\)) and sulfate ions (\(\mathrm{SO}_4^{2-}\)).

The key to calculations involving dissociation lies in the stoichiometry of the compound. For every one mole of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), there will be two moles of \(\mathrm{Al}^{3+}\) and three moles of \(\mathrm{SO}_4^{2-}\). This ratio is critical when calculating the molarity of the individual ions in solution after dissociation occurs. Therefore, understanding the concept of dissociation is a vital step in being able to determine the concentration of each ion in an aqueous solution.

Concentration of Ions in Solution

The concentration of ions in a solution is an expression of the amount of those ions present in a given volume of the solution. One common way to express this concentration is molarity, which is moles of solute per liter of solution (M or mol/L).

In a solution where an ionic compound has dissociated, the concentration of the individual ions will depend on both the molarity of the original compound and the stoichiometry of the dissociation. Since each compound can produce different amounts of ions depending on its formula, just knowing the molarity of the original compound isn't enough; you also need to consider how many ions are produced per formula unit of the compound. This understanding allows chemists to manipulate and utilize solutions with precision in both laboratory and industrial settings.