Problem 82 - Chemistry The Molecular Science

Question

Assume that \(6.73 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is dissolved in enough water to make \(250 . \mathrm{mL}\) solution. (a) Calculate the molarity of the sodium carbonate. (b) Calculate the concentrations of the \(\mathrm{Na}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) ions.

Step-by-Step Solution

Answer

Molarity of Na2CO3 is 0.254 M. Concentrations: Na+ = 0.508 M, CO3^{2-} = 0.254 M.

Step 1: Calculate the Molar Mass of Na2CO3

To find the molarity of the solution, we first need the molar mass of sodium carbonate (Na2CO3). Using the atomic masses: \(\text{Na} = 22.99\;\text{g/mol}\), \(\text{C} = 12.01 \;\text{g/mol}\), \(\text{O} = 16.00 \;\text{g/mol}\), the molar mass of Na2CO3 is calculated as \((2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99\;\text{g/mol}\).

Step 2: Calculate the Moles of Na2CO3

Using the mass and molar mass, calculate the moles of Na2CO3: \(\text{moles Na}_2\text{CO}_3 = \frac{6.73\;\text{g}}{105.99\;\text{g/mol}} \approx 0.0635\;\text{mol}\).

Step 3: Calculate the Molarity of Na2CO3

To calculate the molarity, use the formula \(\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\). Since the solution volume is 250 mL, convert to liters: \(250\;\text{mL} = 0.250\;\text{L}\). Thus, molarity \(= \frac{0.0635\;\text{mol}}{0.250\;\text{L}} = 0.254\;\text{M}\).

Step 4: Use Stoichiometry to Find Ion Concentrations

Sodium carbonate dissolves in water to form ions: \(\text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-}\). Each mole of Na2CO3 produces 2 moles of Na+ and 1 mole of CO3^{2-}.

Step 5: Calculate Concentration of Na+ Ions

The concentration of Na+ ions is twice that of Na2CO3 because each formula unit produces 2 Na+ ions. Thus, \([\text{Na}^+] = 2 \times 0.254\;\text{M} = 0.508\;\text{M}\).

Step 6: Calculate Concentration of CO3^{2-} Ions

The concentration of CO3^{2-} ions is the same as the concentration of Na2CO3. Therefore, \([\text{CO}_3^{2-}] = 0.254\;\text{M}\).

Key Concepts

Molar Mass

To calculate the molarity of a solution, knowing the molar mass of a compound is essential. The **molar mass** is the mass of one mole of a substance, measured in grams per mole (g/mol). To find it, you add up the atomic masses of all the atoms in a molecule.
For sodium carbonate, \( ext{Na}_2 ext{CO}_3\), you need the atomic masses of sodium (Na), carbon (C), and oxygen (O):

• Sodium: 22.99 g/mol
• Carbon: 12.01 g/mol
• Oxygen: 16.00 g/mol

By multiplying the atomic masses by the number of each type of atom in the molecule, you get:\[(2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \text{ g/mol}\]This gives the molar mass of sodium carbonate, which is used to convert grams of a substance to moles, a crucial step in molarity calculation.

Ion Concentration

When ionic compounds like sodium carbonate dissolve in water, they dissociate into individual ions. Understanding **ion concentration** is key to predicting reactions in solutions.
Sodium carbonate, \( ext{Na}_2 ext{CO}_3\), dissociates into sodium ions and carbonate ions:

• \(2\;\text{Na}^+\)
• \(\text{CO}_3^{2-}\)

For every mole of \( ext{Na}_2 ext{CO}_3\) that dissolves, 2 moles of \(\text{Na}^+\) and 1 mole of \(\text{CO}_3^{2-}\) are formed.
The concentration of ions in a solution can be determined based on this stoichiometry:

• \([\text{Na}^+] = 2 \times \text{Molarity of } \text{Na}_2\text{CO}_3 = 0.508\;\text{M}\)
• \([\text{CO}_3^{2-}] = \text{Molarity of } \text{Na}_2\text{CO}_3 = 0.254\;\text{M}\)

This clear relationship helps predict how a solution behaves in chemical reactions.

Stoichiometry

**Stoichiometry** is the study of the quantitative relationships in chemical reactions. It helps us understand how substances react and in what proportions.

In our example, the stoichiometric reaction for sodium carbonate in water is:\[\text{Na}_2\text{CO}_3 \rightarrow 2\;\text{Na}^+ + \text{CO}_3^{2-}\]This equation tells us that one mole of sodium carbonate produces two moles of sodium ions and one mole of carbonate ions. This ratio is essential in calculating ion concentrations from the molarity of a compound.
With this balanced equation, you can see:

• Twice as many \(\text{Na}^+\) ions as sodium carbonate molecules.
• The concentration of \(\text{CO}_3^{2-}\) ions equals the molarity of \(\text{Na}_2\text{CO}_3\).

Understanding stoichiometry ensures accurate calculations and predictions in chemistry, allowing us to control and manipulate reactions efficiently.