Question

If you want a solution that is 0.100 \(m\) in ions, what mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) must you dissolve in \(125 \mathrm{g}\) of water? (Assume total dissociation of the ionic solid.

Short answer

Dissolve 0.5917 g of Na2SO4 in 125 g of water.

Step-by-step solution

Calculate Moles of Ions

Since 1 mole of \( \text{Na}_2\text{SO}_4 \) dissociates into 3 moles of ions (2 \(\text{Na}^+\) ions and 1 \(\text{SO}_4^{2-} \) ion), we need to find the number of moles of ions needed. For a 0.100 \( m \) solution in terms of ions in 125 g water, the moles of ions required is \( \text{moles} = 0.100 \, \text{m} \times 0.125 \, \text{kg} = 0.0125 \, \text{moles of ions} \).

Relate Moles of Ions to Moles of Na2SO4

Since each mole of \( \text{Na}_2\text{SO}_4 \) produces 3 moles of ions, the moles of \( \text{Na}_2\text{SO}_4 \) needed is \( \frac{0.0125}{3} = 0.004167\,\text{moles} \).

Calculate Mass of Na2SO4

Find the molar mass of \( \text{Na}_2\text{SO}_4 \) by adding the atomic masses: \( 2 \times 23 + 32 + 4 \times 16 = 142 \, \text{g/mol} \). Compute the mass required: \( 0.004167 \, \text{moles} \times 142 \, \text{g/mol} = 0.5917 \, \text{g} \).

Key concepts

Ionic Dissociation

When a compound like sodium sulfate (\(\text{Na}_2\text{SO}_4\)) is added to water, it undergoes a process called ionic dissociation. This process involves breaking down the compound into its constituent ions. Understanding ionic dissociation is crucial as it helps determine the concentration of ions in a solution.For example:- \(\text{Na}_2\text{SO}_4\) dissociates into two \(\text{Na}^{+}\) ions and one \(\text{SO}_4^{2-}\) ion.- This means that a single mole of \(\text{Na}_2\text{SO}_4\) produces a total of three moles of ions.This complete dissociation is essential in solving problems related to ion concentration in solutions. Knowing how many ions are produced helps us calculate the molality, which refers to the moles of solute (in this case, ions) per kilogram of solvent (water). Understanding these fundamental details can make balancing chemical equations and predicting the properties of solutions more intuitive.

Moles Calculation

The concept of moles is fundamental in chemistry, allowing us to quantify substances. In this problem, we want a solution with a specific ion concentration. To achieve this, we first calculate the moles of ions required:- The desired concentration is 0.100 \(m\), meaning 0.100 moles of ions per kg of water.- For 125 g of water (0.125 kg), we multiply to find 0.0125 moles of ions needed.Next, we relate this back to \(\text{Na}_2\text{SO}_4\):- Since each mole of \(\text{Na}_2\text{SO}_4\) produces three moles of ions, we divide the total moles of ions by three.- This simple division gives us the moles of \(\text{Na}_2\text{SO}_4\) required: \(\frac{0.0125}{3} = 0.004167\) moles.Calculating moles involves basic arithmetic but deeply ties into understanding how chemical reactions occur at the molecular level. Knowing this calculation also helps when dealing with larger-scale reactions or when preparing specific chemical solutions.

Molar Mass

Molar mass is the weight of one mole of a substance. Knowing this helps in converting between the mass and the amount (in moles) of a substance. For \(\text{Na}_2\text{SO}_4\), its molar mass calculation is based on the atomic masses of its constituent elements:- Sodium (Na): Atomic mass = 23 g/mol. With 2 sodium atoms, it contributes \(2 \times 23 = 46\) g/mol.- Sulfur (S): Atomic mass = 32 g/mol. So, sulfur adds 32 g/mol.- Oxygen (O): Atomic mass = 16 g/mol. With 4 oxygen atoms, it adds \(4 \times 16 = 64\) g/mol.Adding these gives the molar mass of \(\text{Na}_2\text{SO}_4\):\(46 + 32 + 64 = 142\) g/mol.This calculation is critical for converting the moles calculated earlier into grams, which is practical when measuring out the chemical for solutions. For our 0.004167 moles of \(\text{Na}_2\text{SO}_4\):- The mass needed is \(0.004167 \times 142 \approx 0.5917\) g.Understanding molar mass allows you to transition between the microscopic world of molecules and the macroscopic world where we weigh substances.