Problem 78 - Chemistry Matter and Change

Question

How many milliliters of 3.0\(M\) phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) can be made from 95 \(\mathrm{mL}\) of a 5.0 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution?

Step-by-Step Solution

Answer

158 mL of 3.0 M phosphoric acid can be made.

Step 1: Understanding the Concept of Molarity

Molarity (\(M\)) is defined as moles of solute per liter of solution. In dilution problems, the moles of solute remain the same before and after dilution.

Step 2: Identify the Data Given

We have an initial solution with a concentration of 5.0 \(M\) and a volume of 95 \(mL\). We need to find the final volume when the concentration is diluted to 3.0 \(M\).

Step 3: Set Up the Dilution Equation

The formula for dilution is: \( M_1 V_1 = M_2 V_2 \), where \(M_1\) and \(V_1\) are the initial molarity and volume, and \(M_2\) and \(V_2\) are the final molarity and volume.

Step 4: Substitute the Known Values

Substitute the given values into the dilution equation: \( (5.0 \ M)(95 \ mL) = (3.0 \ M)(V_2) \).

Step 5: Solving for the Unknown

Solve for \(V_2\): \[ V_2 = \frac{(5.0 \ M) \times (95 \ mL)}{3.0 \ M} \].

Step 6: Calculate the Final Volume

Perform the calculation: \( V_2 = \frac{475}{3} \approx 158.33 \ mL \). The solution must be expressed to three significant figures.

Key Concepts

Understanding Molarity

Molarity is a fundamental concept in chemistry that refers to the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution, symbolized as

• The formula for molarity is \( M = \frac{n}{V} \), where \( n \) is the number of moles of solute and \( V \) is the volume of the solution in liters.
• Molarity provides a convenient way to express concentration for reactions that occur in solutions.
• In scenarios involving dilution, the key takeaway is that the moles of solute remain unchanged.
• This means any change in molarity due to dilution or concentration is balanced by a change in the solution volume.

For example, when discussing a molarity of 3.0 \(M\) phosphoric acid, it means there are 3 moles of phosphoric acid in every liter of the solution. This consistency enables chemists to predict how much of a solution is needed when changing concentrations or mixing chemicals.

The Role of Phosphoric Acid in Chemistry

Phosphoric acid, denoted as \(H_3PO_4\), is a versatile acid with various applications in both industrial and laboratory settings.

• It has a triprotic nature, meaning it can donate three protons, which makes it a significant player in buffering solutions.
• Commonly used in fertilizers, food flavoring, and cleaning products, its widespread use in labs is due to its stability and relatively weak acidity compared to other mineral acids.
• For students working with phosphoric acid, it's essential to understand its dissociation process in water.
• The dissociation into hydrogen ions and phosphate ions contributes to its acid behavior: \( H_3PO_4 \leftrightarrow H_2PO_4^- + H^+ \).

When considering a dilution problem involving phosphoric acid, like the original exercise, the focus is on adjusting the concentration while the total amount of phosphoric acid (number of moles) remains constant.

Conceptualizing Concentration in Solutions

In chemistry, concentration refers to the amount of solute present in a given quantity of solvent or solution.

• There are various ways to express concentration, such as molarity, molality, and percentage composition.
• Molarity is among the most commonly used, especially in laboratory settings, because it links solute quantity to solution volume directly.
• Understanding concentration is crucial when conducting reactions that require precise quantities of reactants.
• A dilution involves lowering the concentration of a solution by adding more solvent, while keeping the amount of solute unchanged.

For example, in the original exercise, to dilute a 5.0 \(M\) phosphoric acid solution to 3.0 \(M\), the concentration change is achieved by increasing the volume of the solution without changing the number of moles of phosphoric acid present, which is calculated using the equation \( M_1V_1 = M_2V_2 \). This highlights the principle of conservation of mass where the mass of solute stays consistent before and after dilution.