Problem 154 - General Chemistry

Question

Determine the volume of sodium hydroxide solution needed to prepare \(26.2 \mathrm{~g}\) sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), by the reaction \(3 \mathrm{NaOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)\) The sodium hydroxide solution, whose density is \(1.133 \mathrm{~g} / \mathrm{mL}\), contains \(12.0 \% \mathrm{NaOH}\) by mass.

Step-by-Step Solution

Answer

You need 141.0 mL of NaOH solution.

Step 1: Find Molar Mass of Sodium Phosphate

To prepare sodium phosphate (\(\mathrm{Na}_3\mathrm{PO}_4\)), we first need to calculate its molar mass. Sodium (\(\mathrm{Na}\)) has an atomic mass of approximately 23.0 g/mol, phosphorus (\(\mathrm{P}\)) is about 30.97 g/mol, and oxygen (\(\mathrm{O}\)) is about 16.0 g/mol. So the molar mass of \(\mathrm{Na}_3\mathrm{PO}_4\) is \((3\times23.0) + 30.97 + (4\times16.0) = 163.94\) g/mol.

Step 2: Calculate Moles of Sodium Phosphate Needed

We need to find out how many moles of sodium phosphate correspond to the given mass of 26.2 g. The number of moles (\(n\)) is calculated using the formula \(n = \frac{\text{mass}}{\text{molar mass}}\). Thus, \(n = \frac{26.2}{163.94} = 0.1598\) moles of \(\mathrm{Na}_3\mathrm{PO}_4\).

Step 3: Determine Moles of Sodium Hydroxide Required

Using the balanced chemical equation \(3\mathrm{NaOH} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Na}_3\mathrm{PO}_4 + 3\mathrm{H}_2\mathrm{O}\), we see that 3 moles of \(\mathrm{NaOH}\) are needed to produce 1 mole of \(\mathrm{Na}_3\mathrm{PO}_4\). Hence, the moles of \(\mathrm{NaOH}\) required are \(3 \times 0.1598 = 0.4794\) moles.

Step 4: Find Mass of Sodium Hydroxide Required

The molar mass of \(\mathrm{NaOH}\) is calculated as \(23.0 + 16.0 + 1.0 = 40.0\) g/mol. Thus, the mass of \(\mathrm{NaOH}\) required is \(0.4794 \times 40.0 = 19.176\) g.

Step 5: Calculate Volume of NaOH Solution Needed

The sodium hydroxide solution contains 12% \(\mathrm{NaOH}\) by mass. Therefore, the mass of the solution is \(\frac{19.176}{0.12} = 159.8\) g. Then, using the density of the solution (1.133 g/mL), the volume needed is \(\frac{159.8}{1.133} = 141.0\) mL.

Key Concepts

Molar Mass Calculation

Understanding molar mass calculation is key in chemistry as it connects mass with moles, which are fundamental units in chemical reactions. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For any compound, it is calculated by adding the atomic masses of all the atoms present in the molecule. For example, let's consider sodium phosphate (\(\mathrm{Na}_3\mathrm{PO}_4\)). To find its molar mass:

• Sodium (\(\mathrm{Na}\)): 3 atoms, each with an atomic mass of 23.0 g/mol, contributing \(3 \times 23.0\).
• Phosphorus (\(\mathrm{P}\)): 1 atom with an atomic mass of 30.97 g/mol.
• Oxygen (\(\mathrm{O}\)): 4 atoms, each with an atomic mass of 16.0 g/mol, contributing \(4 \times 16.0\).

So, the molar mass of sodium phosphate is calculated as \((3 \times 23.0) + 30.97 + (4 \times 16.0) = 163.94\) g/mol.
This means one mole of sodium phosphate weighs 163.94 grams. Knowing how to calculate molar mass helps predict how much substance is needed or produced in a reaction.

Chemical Reactions

Chemical reactions describe how substances interact to form new substances. A balanced chemical equation provides insights into the reactants and products and their quantities. For our reaction:\[3 \mathrm{NaOH}(aq) + \mathrm{H}_3\mathrm{PO}_4(aq) \rightarrow \mathrm{Na}_3\mathrm{PO}_4(aq) + 3 \mathrm{H}_2\mathrm{O}(l)\]Here, three moles of sodium hydroxide react with one mole of phosphoric acid to produce one mole of sodium phosphate and three moles of water. Balancing the equation ensures that the same number of each type of atom is present on both sides, adhering to the law of conservation of mass.
This particular reaction illustrates a synthesis process, where smaller reactants combine to form a larger product. Understanding chemical reactions helps in calculating the amounts of reactants needed and predicting the quantities of products formed.

Solution Density

Solution density is a measure of how much mass is contained in a given volume of a solution. It is expressed in grams per milliliter (g/mL) and plays a crucial role in mass-volume conversions. In our problem, the density of the sodium hydroxide solution is given as 1.133 g/mL.
Density helps determine the relationship between the mass and volume of a solution. Specifically, it allows us to convert mass into volume by using the formula: \[\text{Volume} = \frac{\text{Mass}}{\text{Density}}\]Using this relationship, once we know the mass of the solution needed (such as from the percentage composition and mass of solute), we can easily determine the volume of the solution that needs to be measured out. This conversion is essential in laboratory settings where precise amounts of substances are required.

Mass-Volume Conversion

Mass-volume conversion is essential in solutions where concentrations are given by percent mass and densities are available. In our scenario, the sodium hydroxide solution is 12% by mass. This means that 12 grams of NaOH are present in every 100 grams of solution.
To find how much solution we need for a specific mass of solute, the formula:\[\text{Mass of solution} = \frac{\text{Mass of solute}}{\text{Percentage as decimal}}\]is used. Once the mass of the solution is known, converting it to volume using solution density becomes straightforward. By dividing the mass of the solution by its density, we get the volume:\[\text{Volume} = \frac{\text{Mass of solution}}{\text{Density}}\]This method allows chemists to accurately measure the amount of liquid solution required to obtain a desired quantity of solute, crucial for reactions and dilutions.