Question

A \(5.00 \mathrm{mL}\) sample of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) requires \(49.1 \mathrm{mL}\) of \(0.217 \mathrm{M} \mathrm{NaOH}\) to convert all of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) to \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) The other product of the reaction is water. Calculate the molarity of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution.

Short answer

The molarity of the Na\(_{2}\)H\(_{3}\)PO\(_{4}\) solution is 1.065 M.

Step-by-step solution

Write and Balance the Chemical Equation

First, we need to correctly write the chemical equation for the reaction and balance it: \[2 \mathrm{NaOH}+\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Na}_{3} \mathrm{PO}_{4}\] The coefficients tell us that 2 moles of NaOH reacts with 1 mole of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) to produce 2 moles of water and 1 mole of \(\mathrm{Na}_{3} \mathrm{PO}_{4}.\)

Calculate the number of moles of NaOH

We can calculate the number of moles of NaOH using its molarity and volume. The molarity (M) is the number of moles of solute per liter of solution, so \[Moles \, of \, NaOH = Molarity \, of \, NaOH \times Volume\] \[Moles \, of \, NaOH = 0.217M \times 49.1mL = 0.0219 \, moles\] We need to convert milliliters to liters by dividing by 1000, so \(Moles \, of \, NaOH = 0.0219 moles \times \frac{1L}{1000mL} = 0.01065 mole\)

Apply the stoichiometric relationships

From the balanced chemical equation, it is clear that 2 moles of NaOH react with 1 mole of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\). Therefore, the number of moles of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) in the reaction would be half of the number of moles of NaOH: \[Moles \, of \, \mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4} = \frac{0.01065}{2} = 0.005325 moles\]

Calculate the Molarity of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\)

We can calculate the molarity of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) using the formula: \[Molarity = \frac{Moles \, of \, solute}{Volume \, of \, solution} = \frac{0.005325 \, moles}{5.00 \, mL} \] Again, we need to convert milliliters to liters by dividing by 1000: \[Molarity = \frac{0.005325 \, moles}{5.00 \, mL/1000} = 1.065 \, M\] Thus, the molarity of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution is 1.065 M.

Key concepts

Molarity Calculation

Molarity, often denoted by the letter \( M \), is a measure of the concentration of a solute in a solution. It is calculated by determining the number of moles of solute in one liter of solution. This concept is essential in chemistry to quantify the concentration of a given chemical when it is dissolved in a solvent.
When calculating molarity, remember the basic formula:

• \( \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \)

In our exercise, we began by finding out the number of moles of NaOH using its given molarity and volume. Afterward, based on the balanced equation, we determined the number of moles of NaOH involved in the reaction. Finally, we used this information to find the molarity of Na₂H₃PO₄ by dividing the moles of solute by the volume of the solution (which was converted from milliliters to liters).
Remember to always convert all measurements to appropriate units, especially when dealing with milliliters to liters, because molarity calculations rely heavily on units being consistent across the calculation process.

Balanced Chemical Equations

Balanced chemical equations represent a fundamental aspect of chemical stoichiometry. These equations display the reactants and products involved in a chemical reaction, with coefficients indicating the proportion of each substance.
In the given exercise, we have the chemical reaction:

• \( 2 \text{NaOH} + \text{Na}_2\text{H}_3\text{PO}_4 \rightarrow 2 \text{H}_2\text{O} + \text{Na}_3\text{PO}_4 \)

This equation signifies that two moles of sodium hydroxide (NaOH) react with one mole of sodium hydrogen phosphate (Na₂H₃PO₄) to produce two moles of water (H₂O) and one mole of sodium phosphate (Na₃PO₄). Understanding this equation is crucial because it allows us to determine the stoichiometric relationships which help in calculating the moles needed for our molarity calculations.
To balance an equation, one must ensure the number of each type of atom on the reactant side equals the number on the product side. This respects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Balanced equations serve as the blueprint for reacting substances in proper proportions.

Acid-Base Titration

Acid-base titration is a quantitative chemical analysis method used to determine the concentration of an acid or base in a solution. During titration, a solution of known concentration, called the titrant, is gradually added to a measured amount of the analyte, which is the solution of unknown concentration.
In our exercise, NaOH is the titrant, and Na₂H₃PO₄ is the analyte. The point at which the reaction is complete is known as the equivalence point and can be identified by observing changes in the solution through the use of indicators or pH measurements.
Titrations require balanced chemical equations to determine the stoichiometric relationship between the titrant and the analyte. Here, we understood from the equation that every two moles of NaOH reacts with one mole of Na₂H₃PO₄, allowing us to precisely calculate the moles and hence the concentration of the unknown solution.
This method is powerful because it provides a precise and accurate measurement of concentration, which is helpful in various fields, such as quality control, environmental testing, and pharmaceuticals.