Problem 40 - Chemistry and Chemical Reactivity

Question

What is the mass of solute, in grams, in 125 mL. of a \(1.023 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\) What is the molar concentration of the \(\mathrm{Na}^{+}\) and \(\mathrm{PO}_{4}^{3-}\) ion?

Step-by-Step Solution

Answer

Mass of solute: 0.02095 g; \(\mathrm{Na}^{+}\) concentration: 3.069 \times 10^{-3} \ \mathrm{M}; \(\mathrm{PO}_{4}^{3-}\) concentration: 1.023 \times 10^{-3} \ \mathrm{M}.

Step 1: Calculate Moles of Solute

First, use the molarity formula \( M = \frac{n}{V} \) to calculate the moles of solute, \( n \). The molarity \( M \) is given as \( 1.023 \times 10^{-3} \ \mathrm{mol/L} \) and the volume \( V \) is 125 mL or 0.125 L.\[ n = M \times V = (1.023 \times 10^{-3} \ \mathrm{mol/L}) \times (0.125 \ \mathrm{L}) = 1.27875 \times 10^{-4} \ \mathrm{mol} \]

Step 2: Convert Moles to Grams

Now we convert the moles of \( \mathrm{Na}_{3} \mathrm{PO}_{4} \) to grams. Use its molar mass, which is approximately 163.94 \( \mathrm{g/mol} \).\[ \text{mass} = n \times \text{molar mass} = (1.27875 \times 10^{-4} \ \mathrm{mol}) \times (163.94 \ \mathrm{g/mol}) = 0.02095 \ \mathrm{g} \]

Step 3: Determine Concentration of Ions

\( \mathrm{Na}_{3} \mathrm{PO}_{4} \) dissociates into 3 \( \mathrm{Na}^{+} \) ions and 1 \( \mathrm{PO}_{4}^{3-} \) ion per formula unit. Use the initial molarity of the solution to find the concentrations of each ion.- For \( \mathrm{Na}^{+} \):\[ 3 \times 1.023 \times 10^{-3} \ \mathrm{M} = 3.069 \times 10^{-3} \ \mathrm{M} \]- For \( \mathrm{PO}_{4}^{3-} \):\[ 1.023 \times 10^{-3} \ \mathrm{M} \]

Key Concepts

Molar mass

To calculate the mass of a solute in grams, understanding the concept of molar mass is crucial. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It allows us to convert between the amount of substance (in moles) and its mass (in grams).
For example, when determining the mass of a solution of sodium phosphate, \(\mathrm{Na}_3\mathrm{PO}_4\), you must first know its molar mass, which is calculated by summing the atomic masses of all atoms in the compound:

• Sodium (Na): 22.99 g/mol × 3 = 68.97 g/mol
• Phosphorus (P): 30.97 g/mol
• Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Adding these up, you find that the molar mass of \(\mathrm{Na}_3\mathrm{PO}_4\) is approximately 163.94 g/mol.Once you have the moles of \(\mathrm{Na}_3\mathrm{PO}_4\) (as calculated from molarity and volume), you can convert to grams using this molar mass.
Just multiply the number of moles by the molar mass to find the mass in grams.This is an essential step in determining how much of a substance is present in a given volume of solution.

Ion concentration

Ion concentration is a measure of the number of ions present in a unit volume of solution. It is typically expressed in moles per liter (M), also known as molarity.
When you dissolve a compound like \(\mathrm{Na}_3\mathrm{PO}_4\) in water, the compound dissociates into its constituent ions. In this case:

• Three \(\mathrm{Na}^{+}\) ions for each \(\mathrm{Na}_3\mathrm{PO}_4\) formula unit
• One \(\mathrm{PO}_4^{3-}\) ion for each \(\mathrm{Na}_3\mathrm{PO}_4\) formula unit

The initial molarity of the solution gives the concentration of the entire compound.To find the concentration of individual ions, you need to multiply this value by the number of each type of ion produced from one formula unit. For instance, for \(\mathrm{Na}^{+}\) ions, multiply the molarity of \(\mathrm{Na}_3\mathrm{PO}_4\) by three, and for \(\mathrm{PO}_4^{3-}\) ions, just use the original molarity. This way, you can accurately assess the ratio of ions in solutions, crucial for various applications like solution preparation and analysis.

Chemical dissociation

Chemical dissociation refers to the process by which a compound splits into its constituent ions in solution. Understanding dissociation is vital for predicting the behavior of compounds in aqueous solutions.
For ionic compounds like \(\mathrm{Na}_3\mathrm{PO}_4\), they dissociate into their respective ions when dissolved in water. This process is central to understanding how these compounds contribute to the composition and properties of solutions.

• \(\mathrm{Na}_3\mathrm{PO}_4 (aq) \rightarrow 3 \mathrm{Na}^{+} (aq) + \mathrm{PO}_4^{3-} (aq)\)

Each formula unit of \(\mathrm{Na}_3\mathrm{PO}_4\) yields three sodium ions and one phosphate ion. Recognizing dissociation patterns allows for the calculation of individual ion concentrations from a known concentration of the parent compound.Notably, dissociation is specific not just to sodium phosphate but to many other ionic compounds used in various fields. Knowing how to calculate resulting ion concentrations is a key skill in chemistry, aiding in solutions' preparation, reactions, and applications in real-world scenarios.